B    M    EMT   bflM 


IN   MEMORIAM 
FLOR1AN  CAJORI 


X 


if- 


PLANE  AND  SOLID 


ANALYTIC    GEOMETRY 


AN  ELEMENTARY  TEXT-BOOK 


BY 


CHARLES   H.    ASHTON,  A.M. 

ASSISTANT   PROFESSOR   OF   MATHEMATICS   IN   THE   UNIVERSITY 
OF    KANSAS 


RE  riSED  ~  EDITION 


-  '  ' 


NEW   YORK 
CHARLES   SCRIBNER'S   SONS 

1904 


Wof 


COPYRIGHT,   1900,   BY 
CHARLES  SCRIBNER'S  SONS 


CAJORI 


Norruoofc  $rrss 

J.  S.  Cushing  &  Co.  —  Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


PREFACE 

The  present  work  is  intended  as  a  text-book  for  class- 
room use,  and  not  as  an  exhaustive  treatise  on  the  sub- 
ject. This  object  has  been  kept  constantly  in  mind  in 
writing  the  book,  and  every  subject  has  been  treated  from 
this  point  of  view.  A  large  part  of  the  book  was  mimeo- 
graphed and  tested  by  use  for  several  years  in  the  author's 
classes  in  Harvard  University. 

The  author  has  tried  to  meet  the  needs  of  a  class  which 
occupies  from  sixty  to  seventy  recitation  hours  upon  the 
subject,  and  it  is  thought  that  the  book  ought  to  be  com- 
pleted by  the  average  class  in  that  time.  Necessarily 
some  subjects  which  usually  find  a  place  in  books  on 
Analytic  Geometry  have  been  omitted ;  but  it  is  thought 
that  nothing  has  been  omitted  which  has  an  important 
bearing  on  future  mathematical  study. 

The  conies  have  been  treated  from  their  ratio  defini- 
tion, and  much  space  and  time  have  been  gained  by  not 
repeating  proofs  which  are  identical,  or  very  similar,  for 
the  three  forms  of  the  conic.  Analytic  methods  are  used 
throughout  the  book,  and  the  author  has  attempted  to 
give  proofs  which  are  concise  and  easily  understood  by 
the  average  student,  but,  at  the  same  time,  mathemati- 
cally rigorous.  In  this  connection  he  would  call  atten- 
tion to  the  proofs  in  oblique  coordinates  (Arts.  12,  28), 
which  are  usually  given  without  reference  to  the  direc- 
tions of  the  lines,  and,  therefore,  do  not  hold  if  the 
positions  of  the  points  are  changed. 

v 

911318 


yi  PREFACE 

Numerous  problems,  which  have  been  selected  with 
great  care,  have  been  inserted  after  nearly  every  article. 
In  the  early  part  of  the  book  these  are  mainly  numerical, 
but  later  the  student  has  been  asked  to  prove  a  large 
number  of  theorems.  A  considerable  number  of  theorems 
which  are  usually  proved  in  the  text  are  here  inserted 
as  problems,  and  in  many  places  the  student  has  been 
asked  to  derive  formulas  for  two  of  the  conies,  after  the 
corresponding  formulas  for  one  of  the  conies  have  been 
obtained.  It  is  only  by  solving  such  problems  that  the 
student  can  acquire  any  real  grasp  of  the  subject. 

The  attention  of  the  teacher  is  also  called  to  the  two 
chapters  on  loci  (Chaps.  VIII  and  XIV),  in  which  a  large 
number  of  problems  are  given  and  the  methods  of  solving 
them  discussed ;  to  the  treatment  of  poles  and  polars  by 
the  aid  of  harmonic  division  (Chap.  XII);  and  to  the 
system  of  polar  coordinates  used  in  the  Solid  Geometry. 

The  author  desires  to  acknowledge  gratefully  the  assist- 
ance of  Mr.  B.  E.  Carter  of  the  Massachusetts  Institute 
of  Technology,  who  has  read  with  great  care  both  manu- 
script and  proof  ;  of  Mr.  E.  V.  Huntington,  who  made 
many  valuable  suggestions  during  the  early  stages  of 
the  work  ;  and  of  Mr.  W.  R.  Marsh,  his  colleague  in  the 
preparation  of  the  series,  of  which  this  volume  is  the  first 
to  appear. 


Cambridge, 

November,  1900. 


CONTENTS 

PART   I 

PLANE  ANALYTIC  GEOMETRY 

CHAPTER  I 

Introduction 

ART.  PAGE 

1.  Directed  lines 1 

2.  Addition  of  directed  lines 1 

3.  Directed  angles 2 

4.  Addition  of  directed  angles 2 

5.  Measurement  of  lines  and  angles 3 

6.  Angles  between  two  lines 4 

7.  Law  of  sines  and  law  of  cosines 5 

8.  The  quadratic  equation 6 

CHAPTER  II 
The  Point 

9.  Cartesian  coordinate  systems 7 

10.  Notation 11 

11.  Distance  between  two  points  in  rectangular  coordinates        .  11 

12.  Distance  between  two  points  in  oblique  coordinates      .         .  13 

13.  Points  dividing  a  line  in  a  given  ratio 14 

14.  Harmonic  division  .        . 16 

CHAPTER  III 
Loci 

15.  Equation  of  a  locus 20 

16.  Locus  of  an  equation 24 

17.  Plotting  the  locus  of  an  equation 25 

vii 


Vlll 


CONTENTS 


ART.  PAGE 

18.  Symmetry 27 

19.  Intercepts 31 

20.  Intersection  of  two  curves 31 

21.  Locus  of  u  +  kv  =  0,  and  uv  =  0 31 


CHAPTER  IV 


The  Straight  Line 


22.  Introduction    .         . 

23.  Line  through  two  points 

24.  Line  determined  by  its  intercepts  .... 

25.  Oblique  coordinates 

26.  Line  determined  by  a  point  and  its  direction 

27.  Line  determined  by  its  slope  and  its  intercept  on  the  ' 

28.  Oblique  coordinates 

29.  General  equation  of  the  first  degree 

30.  Two  equations   representing  the  same  line  cannot 

except  by  a  constant  factor         .         .         ... 

31.  The  angle  which  one  line  makes  with  another 

32.  Perpendicular  and  parallel  lines     .... 

33.  Line  making  a  given  angle  with  a  given  line 

34.  Normal  form  of  the  equation  of  a  straight  line 

35.  Reduction  of  the  general  equation  to  the  normal  form 

36.  Distance  of  a  point  from  a  line       .... 

37.  Oblique  coordinates 

38.  Bisector  of  the  angle  between  two  lines 

39.  Lines  through  the  intersection  of  two  given  lines  . 

40.  Area  of  a  triangle 


differ 


■axis 


34 
34 
36 
36 

37 
38 
39 
40 

42 
44 
46 
48 
49 
51 
53 
55 
56 
57 
59 


CHAPTER  V 

Polar  Coordinates 

41.  Introduction 63 

42.  Equation  of  a  locus 64 

43.  Plotting  in  polar  coordinates 65 

44.  Natural  values  of  the  sines,  cosines,  tangents,  and  cotangents  66 


CONTENTS  IX 


CHAPTER    VI 


Transformation  of  Coordinates 

ART.  PAGE 

45.  Introduction 68 

46.  Transformation  to  axes  parallel  to  the  original  axes     .         .      68 

47.  Transformation  from  one  set  of  rectangular  axes  to  another, 

having  the  same  origin  and  making  an  angle  0  with  the 
first  set 70 

48.  Transformation  in  which  both  the  position  of  the  origin 

and  the  direction  of  the  axes  are  changed  .         .         .         .71 

49.  Transformation  from   any  Cartesian  system  to  any  other 

Cartesian  system,  having  the  same  origin  ....       72 

50.  Degree  of  an  equation  not  changed  by  transformation  of 

coordinates •         •         .72 

51.  Transformation  from  rectangular  to  polar  coordinates  .         .      73 


CHAPTER  VII 
The  Circle 

52.  Equation 75 

53.  General  form  of  the  equation 76 

54.  Circle  through  three  points 77 

55.  Tangent 79 

56.  Normal 81 

57.  Tangents  from  an  exterior  point 82 

58.  Tangent  in  terms  of  its  slope 84 

59.  Chord  of  contact 85 


CHAPTER  VIII 
Loci 

60.  Problems 88 

61.  Problems 91 


CONTENTS 


CHAPTER   IX 
Conic  Sections 

ART.                                                                                                                                    .  PAGE 

62.  Definition  and  equation 100 

63.  Parabola 101 

64.  Central  conies 103 

65.  Ellipse 106 

66.  Hyperbola 110 

67.  Asymptotes 114 

68.  Conjugate  hyperbolas 116 

69.  Equilateral  or  rectangular  hyperbola 117 

70.  Focal  radii  of  a  central  conic 118 

71.  Mechanical  construction  of  the  conies 120 

72.  Auxiliary  circles 121 

73.  General  equation  of  conies  when  axes  are  parallel  to  the 

coordinate  axes 123 


CHAPTER  X 

Tangents 

74.  Equations  of  tangents 126 

75.  Normals 128 

76.  Subtangents  and  subnormals 129 

77.  Slope  form  of  the  equations  of  tangents         ....  131 

78.  Theorems  concerning  tangents  and  normals  ....  133 


CHAPTER  XI 
Diameters 

79.  Equations  of  diameters 142 

80.  Conjugate  diameters 144 

81.  Equation  of  conjugate  diameter 147 

82.  Theorems  concerning  diameters 148 


CONTENTS  XI 


CHAPTER   XII 


Poles  and  Polars 

ART.  PAGE 

83.  Harmonic  division 154 

84.  Polar  of  a  point 155 

85.  Position  of  the  polar 157 

86.  Theorems  concerning  poles  and  polars 159 


CHAPTER  XIII 
General  Equation  of  the  Second  Degree 

87.  Introduction 166 

88.  Two  straight  lines 166 

B*  -  4  .4  C  ^  0. 

89.  Removal  of  the  terms  of  the  first  degree         .         .        .        .168 

90.  Removal  of  the  term  in  ^ 169 

91.  Determination  of  the  coefficients  A',  C,  and  F'     .         .         .  170 

92.  Nature  of  the  locus 172 

B*  -  4  A  C  =  0. 

93.  Removal  of  the  term  in  xy 175 

94.  Removal  of  the  term  my 177 

95.  Nature  of  the  locus 177 

96.  Second  method  of  reducing  the  general  equation  to  a  simple 

form,  when£-4.4C  =  0 178 

97.  Summary 182 

98.  General  equation  in  oblique  coordinates         ....  183 

99.  Conic  through  five  points 183 


CHAPTER  XIV 
Problems  in  Loci •    185 


Xll 


CONTENTS 


CHAPTER  XV 
Higher  Plane  Curves 

ART.  PAGE 

100.  Introduction 195 

101.  The  parabolas 195 

102.  The  Cassinian  oval 196 

103.  The  cissoid 198 

104.  The  conchoid 198 

105.  The  cycloid 200 

106.  The  hypocycloid 201 

107.  The  epicycloid 203 

108.  The  cardioid 204 

109.  The  catenary 205 

110.  The  spirals 205 


PART   II 


ANALYTIC  GEOMETRY  OF  SPACE 


CHAPTER  I 


Coordinate  Systems.     The  Point 


1.  Introductory 

2.  Rectangular  coordinates 

3.  Distance  between  two  points 

4.  To  divide  a  line  in  any  given  ratio 

5.  Projection  of  a  given  line  on  a  given  axis 

6.  Polar  coordinates 

7.  Spherical  coordinates    .... 

8.  Angle  between  two  lines 

9.  Transformation  of  coordinates.     Parallel  axes 
10.  Transformation  of  coordinates  from  one  set  of  rectangular 

axes  to  another  which  has  the  same  origin 


207 
208 
209 
210 
211 
212 
215 
216 
218 

218 


CONTENTS  X11L 

CHAPTER  II 
Loci 

ART.  PAGB 

11.  Equation  of  a  locus 220 

12.  Cylindrical  surfaces 221 

13.  Surfaces  of  revolution 221 

11.   Locus  of  an  equation 223 

CHAPTER.  Ill 
The  Plane 

15.  Normal  form  of  the  equation  of  a  plane        ....  227 

16.  Reduction  of  the  general  equation  A  x  +  By  +  Cz  +  D  =  0  to 

the  normal  form 228 

Equation  of  a  plane  in  terms  of  its  intercepts        .         .         .  229 

Distance  of  a  point  from  a  plane 229 

The  angle  between  two  planes 231 

Perpendicular  and  parallel  planes  .......  232 

Equation  of  a  plane  satisfying  three  conditions     .         .         .  232 

CHAPTER    IV 
The  Straight  Line 

22.  Equations 235 

23.  The  equations  of  a  line  in  terms  of  its  direction  cosines  and 

the  coordinates  of  a  point  through  which  it  passes    .         .  237 

24.  Given  the  equations  of  a  line,  to  find  its  direction  cosines    .  238 

25.  Equations  of  a  line  through  two  points          ....  240 

CHAPTER  V 
Quadric  Surfaces* 

26.  The  sphere 242 

27.  Conicoids 244 

28.  The  ellipsoid  . 245 


XIV 


CONTENTS 


29.  The  imparted  hyperboloid 

30.  The  biparted  hyperboloid 

31.  The  cone 

32.  Asymptotic  cones    . 

33.  The  paraboloids 

34.  Ruled  surfaces 

35.  Tangent  planes 

36.  Normals  . 

37.  Diametrical  planes 

38.  Polar  plane 


PAGE 

247 
250 
252 
253 

254 
257 
259 
261 
262 
264 


Answers 


PART   I 

PLANE  ANALYTIC  GEOMETRY 
CHAPTER   I 

INTRODUCTION 

1.  Directed  lines.  —  If  a  point  moves  from  A  to  B  in 
a  straight  line,  Ave  shall  say  that  it  generates  the  line 
AB ;  if  it  moves  from  B  to  A,  it  generates  the  line  BA. 

In  our  study  of  Geometry,  AB     ^ ^ -c 

and  BA  meant  the  same  thing,  —     j ^ B 


the  line  joining  A  and  B  with-     c j B 

out  regard  to  its  direction.    But  FlG-  1« 

we  shall  now  find  it  convenient  to  distinguish  between  AB 
and  BA  as  if  they  were  separate  lines.  The  position  from 
which  the  generating  point  starts  is  called  the  initial  point 
of  the  line ;  the  point  where  it  stops,  the  terminal  point. 

2.  Addition  of  directed  lines.  —  If  a  point  moves  in  a 
straight  line  from  A  to  B  (on  any  one  of  the  lines  in 
Fig.  1)  and  then  moves  in  that  line,  or  in  that  line  pro- 
duced, to  (7,  the  position  which  it  finally  reaches  is  evi- 
dently the  same  as  if,  starting  from  A,  it  had  moved 
along  the  single  line  AC.  The  line  AC  is  called  the  sum 
of  the  lines  AB  and  BO;  that  is,  AB  +  BO=AO.  Evi- 
dently AB  +  BA  =  AA  =  0,  and  hence  AB  =  -  BA. 

l 


2'  ANALYTIC   GEOMETRY  [Ch.  I,  §§  3, 4 

3.  Directed  angles.  —  If  a  line  starts  from  the  position 
OA  and  rotates  in  a  fixed  plane  about  the  point  0  into 
the  position  OB,  it  is  said  to  generate  the  angle  A  OB. 
If  it  rotates  from  OB  to  OA,  it  generates  the  angle  BOA. 

We  shall  find  it  convenient   to 

^"  ^-^B  distinguish   between  the  angles 

/         ^ ^    /  \  A  OB  and  BOA  as  if  they  were 

/        /  jP^\       f\  separate  angles.      The   position 

\ff       \s       qZ- _J \a  from    which    the    moving    line 

\  /        /  starts  is  called  the  initial  side 

^t ~S        /  of  the  angle  ;  the  position  where 

^S  it  stops,  the  terminal  side. 

There  is  no  limit  to  the  pos- 
Eig    2 

sible  amount  of  rotation  of  the 

moving  line ;  after  performing  a  complete  revolution  in 
either  direction,  it  may  continue  to  rotate  as  many 
times  as  we  please,  generating  angles  of  any  magnitude 
in  either  direction.  Angles  which  are  not  equal,  but 
have  the  same  initial  and  terminal  sides  (1  and  3,  or  2 
and  4,  Fig.  2)  are  called  congruent  angles. 

In  reading  the  angle  AOB,  it  is  not  possible  to  dis- 
tinguish between  the  various  congruent  angles  which 
have  OA  and  OB  for  their  initial  and  terminal  lines,  but 
we  shall  understand  that  the  smallest  of  the  congruent 
angles  is  meant,  unless  another  angle  is  indicated  by  an 
arrow  in  the  figure. 

4.  Addition  of  directed  angles.  —  If  the  moving  line 
starts  from  OA  (in  any  one  of  these  figures)  and  rotates 
first  through  the  angle  A  OB,  and  then  through  the  angle 
BOO,  it  is  evident  that  the  position  00,  which  the  line 


C.i.  I,  §  5] 


INTRODUCTION 


3 


finally  reaches,  is  the  same   as  if,  starting  from    OA,   it 
had  rotated  through  the  single  angle  A  00.     The  angle 


-^J3 


Fig.  3. 

A  00  is  called  the  sum  of  the  angles  A  OB  and  BOO; 
that  is,  ZAOB  +ZB0O=ZA0O.  Evidently  ZAOB  + 
ZBOA  =  0,  and  hence  ZAOB  =  -ZBOA. 

5.  Measurement  of  lines  and  angles.  —  The  length  of 
a  line,  or  the  magnitude  of  an  angle,  may  be  represented 
by  a  number,  by  the  familiar  process  of  measurement. 
That  is,  the  number  of  times  which  the  given  line  or 
angle  contains  an  arbitrarily  chosen  unit  may  be  used 
to  represent  the  length  of  the  line  or  the  magnitude  of 
the  angle.  But  we  have  seen  that  it  is  necessary  to 
distinguish  between  the  lines  AB  and  BA,  and  that  it 
has  followed  from  our  definition  of  addition  of  lines  that 
AB  =  —  BA.  Hence,  if  the  line  AB  is  represented  by  a 
positive  number,  the  line  BA  ivill  be  represented  by  the 
same  number  with  a  negative  sign.  In  like  manner,  if 
the  angle  A  OB  is  represented  by  a  positive  number,  the 
angle  BOA  will  be  represented  by  the  same  number  with 
a  negative  sign.  It  follows,  therefore,  that  opposite  direc- 
tions are  indicated  by  opposite  signs ;  that  is,  if  the  length 
of  a  line  or  the  magnitude  of  an  angle,  generated  in  one  direc- 
tion, is  represented  by  a  positive  number,  then  the  length  of 
a  line  or  the  magnitude  of  an  angle  generated  in  the  opposite 


ANALYTIC   GEOMETRY 


[Ch.  I,  §  6 


direction,  is  represented  by  a  negative  number.  Either  of 
two  opposite  directions  may  be  chosen  as  the  positive 
direction ;  then  the  other  must  be  taken  as  the  negative. 

As  all  our  work  will  be  concerned  with  the  algebraic 
number  rather  than  the  geometric  line  which  it  represents, 
it  will  not  be  necessary  to  distinguish  between  the  line  AB 
and  the  number  which  represents  its  length.  We  shall 
let  AB  stand  for  the  number  which  represents  the  length  of 
the  line  from  A  to  B.  It  is  easily  shown  that  the  length 
of  the  sum  of  two  or  more  lines  that  run  in  the  same  or 
in  opposite  directions  is  the  algebraic  sum  of  the  lengths 
of  the  separate  lines.  Hence  it  is  still  true  that 
AB+BC=  AC,  when  AB,  BO,  and  AC  stand  for  the 
lengths  of  the  lines  AB,  BO,  and  AC.  Since  these  are 
now  algebraic  numbers,  it  follows  that  AB  =  AC—  BO. 

In  like  manner  A  OB  will  be  used  to  represent  the 
magnitude  of  the  angle  instead  of  the  angle  itself.  With 
this  meaning  it  will  still  be  true  that 

ZAOB  +  ZBOC=ZAOC. 

Also,  ZAOB=ZAOC-ZBOC. 

6.  Angles  between  two  lines.  — 
When  two  lines  intersect  at  a  point, 
they  form  several  angles  at  that  point. 
To  avoid  ambiguity,  if  the  lines  are 
directed  lines,  we  shall  define  the 
angle  between  them  as  the  angle  from 
the  positive  direction  of  the  first  line  to 
the  positive  direction  of  the  second  line, 
the  smallest  of  the  congruent  angles 
being  chosen. 


Cn.   I,  §  7]  INTRODUCTION  5 

We  shall  adopt  the  following  notation  :  Denoting  the 
intersecting  lines  by  single  letters,  as  a  and  b,  the  symbol 
(a,  b)  shall  indicate  the  angle  from  the  positive  direction 
of  the  line  a  to  the  positive  direction  of  the  line  6,  to  be 
read,  "the  angle  from  a  to  b." 

It  will  sometimes  be  inconvenient  to  choose  either  direc- 
tion of  an  unlimited  line  as  positive.  (As  when  the  line 
is  given  by  its  equation.)  We  shall  then  define  the  angle 
which  one  line  makes  with  another  as  the  angle  formed  in 
going  from  the  second  to  the  first  in  the  positive  direc- 
tion of  rotation. 

It  is  customary  to  call  the  angle  from  a  to  b  positive  if 
its  rotation  is  opposite  to  that  of  the  hands  of  a  clock;  nega- 
tive if  in  the  same  direction  as  the  hands. 

7.    Law  of  sines  and  law  of  cosines.  —  The    two   laws 
concerning  the  sines  and  the  cosines  of  the  angles  of  a 
triangle  are  often  stated  in  trigonometry  without  regard 
to  the  direction  of  the  sides  of 
the  triangle.     But  for  our  work 
these  must  be  stated  in  a  more 
accurate  form.      Let  the  posi- 
tive   direction  of   each  side   of 
the  triangle  ABC  be  fixed.     It  "  FlG   5 

can  be  easily  shown  that  these 

two  laws  take  the  following  form,  when  the  directions  of 
the  lines  and  angles  are  considered. 

T  p    .  •  AB      sin  (a,  b) 

Law  ot  sines  :  — -  = . 

BO     sin  (6,  c) 

Law  of  cosines  : 
(AB)2  =  (BC)2  +  (CA)2  +  2(BC)(CA)  cos(a,  b). 


6  ANALYTIC   GEOMETRY  [Ch.  I,  §  8 

8.  The  quadratic  equation.  — We  shall  have  occasion  to 
use  a  few  theorems  in  quadratic  equations  which  it  seems 
advisable  to  reproduce  here. 

Any  quadratic  equation  may  be  written  in  the  form 

ax2  -f  bx  +  c  =  0. 
The  two  roots  of  this  equation  are 


2  a 

,  an 

a  x2  = 

By  addition, 

X\  ~1"    r2  ~ 

b 
a 

By  multiplication, 

x^x2  = 

c 
a 

b-Vb2-4:  ac 
2a 


The  sum  and  the  product  of  the  roots  can  therefore  be 
found  directly  from  the  equation  without  solving. 

The  character  of  the  roots  depends  on  the  quantity 
under  the  radical,  b2  —  4  ac. 

If  b2  —  4  ac  >  0,  the  roots  are  real  and  unequal, 

if  b2  —  4  ac  =  0,  the  roots  are  real  and  equal, 

if  b2  —  4  ac  <  0,  the  roots  are  imaginary. 

This  quantity,  b2  —  4  ac,  is  called  the  discriminant  of  the 
equation,  and  when  placed  equal  to  zero  expresses  the 
condition  which  must  hold  between  the  coefficients,  if 
the  two  roots  of  the  equation  are  equal. 


CHAPTER    II 

THE    POINT 

9.  Cartesian  coordinate  systems.  —  The  subject  of  Ana- 
lytic Geometry  is,  as  its  name  implies,  a  treatment  of 
Geometry  by  analytic  or  algebraic  methods.  It  is  then 
essential  to  have  the  means  of  translating  geometric 
statements  into  algebraic  and  the  reverse.  Geometric 
theorems  involve  the  ideas  of  magnitude,  position,  and 
direction.  Algebraic  methods  of  representing  magnitude 
and  direction  have  been  considered  in  the  previous  chapter. 

The  idea  of  position  may  be  expressed  algebraically  in 
many  ways.  But  at  present  we  shall  confine  ourselves  to 
two  methods  used  in  ordinary  life.  If  we  wish  to  locate 
a  town,  Ave  usually  speak  of  it  as  being  a  certain  distance 
in  a  certain  direction  from  some  well-known  location.  In 
the  plane  we  must  have  a  fixed  point  A  from  which  to 
measure  the  distance,  and  a  fixed  line  AB  from  which  to 
measure  the  direction  of  any 
point  P.  The  point  P  is  com- 
pletely determined  when  the 
angle  BAP  and  the  distance 
AP  are  given.  This  system 
of  locating  points  in  a  plane  is  1G' 

called  the  Polar  System,  and  will  be  discussed  fully  later. 

Another  method  of  fixing  the  position  of  a  point  on  the 
earth's  surface  is  to  give  its  latitude  and  longitude,  or  its 

7 


8 


ANALYTIC   GEOMETRY 


[Oh.  II,  §  9 


AY 


IT 


X 


III 


A 


IV 


Y' 
Fig.  7. 


distance  north  or  south  and  its  distance  east  or  west  from 
a  given  pair  of  perpendicular  lines. 

Constructing  a  pair   of   perpendicular  lines  X'X  and 
Y'  Y  in  the  plane,  we  may  locate  a  point  by  saying  that  it 

is  m  units  above  or  below 
X'X  and  n  units  to  the 
right  or  left  of  Y'Y.  If 
instead  of  using  the  words 
above  or  below,  right  or 
left,  we  understand  that 
all  distances  measured  up- 
ivard  or  to  the  right  are 
positive,  and  those  meas- 
ured downward  or  to  the 
left  are  negative,  two  num- 
bers with  the  proper  signs 
attached  will  represent  the  distances  of  the  point  from  the 
two  lines,  and  these  two  numbers  taken  together  will  locate 
absolutely  the  position  of  any  point  in  the  plane.  These 
numbers,  representing  the  distances  of  the  point  from  the 
two  lines,  with  their  proper  signs  attached,  are  called  the 
coordinates  of  the  point.  The  distance  NP,  measured  from 
Y'  Y,  parallel  to  XrX,  is  called  the  abscissa,  or  jr-cobrdinate, 
and  the  distance  MP,  measured  from  X'X,  parallel  to  Y'  Y, 
is  called  the  ordinate,  or  /-coordinate,  of  the  point.  The 
line  X'X  is  called  the  axis  of  abscissas,  or  Jf-axis,  and 
Y'Y  the  axis  of  ordinates,  or  X-axis.  The  two  lines  to- 
gether are  called  the  axes  of  coordinates,  or  coordinate  axes, 
and  their  intersection  the  origin  of  coordinates,  or  simply 
the  origin.  The  abscissa  of  a  point  is  denoted  by  the 
letter  x,  the  ordinate  by  y,  and  the  two  coordinates  are 


oi.  ii,  §  yj  tiie  roiMT  9 

written  in  a  parenthesis  (x,  y),  the  abscissa  being  always 
written  first. 

It  will  be  seen  at  once  that  any  point  in  the  plane  can 
be  located  by  means  of  its  coordinates,  and  that  there  will 
always  be  a  point  which  will  correspond  to  any  pair  of 
values  we  may  choose,  and  that  there  will  be  only  one 
such  point.  We  have  then  a  simple  means  of  representing 
position  in  a  plane  by  algebraic  symbols.  This  system  is 
called  the  rectangular,  and  is  a  particular  case  of  Cartesian 
coordinates.  In  the  general  Cartesian  system  the  axes 
are  not  necessarily  perpendicular  to  each  other.  In  case 
they  are  not  perpendicular,  the  system  is  called  oblique. 
All  the  definitions  given  above  hold  for  the  oblique 
system. 

In  Fig.  8,  NP  is  the  abscissa  of  P  and  MP  is  its  ordi- 
nate. While  rectangular  coordinates  are  more  often  used 
because  their  formulas  are 
simpler,  yet  it  will  occa- 
sionally be  desirable  to  use 
the  more  general  system. 
But  rectangular  coordi- 
nates will  always  be  un- 
derstood unless  another 
system  is  distinctly  speci- 
fied. 

In  locating  or  'plotting  a 
point  whose  coordinates  are  given,  some  convenient  unit 
of  measure  must  first  be  chosen.  Then  measure  off  the 
proper  number  of  these  units  from  the  origin  along  each 
axis  in  the  direction  indicated  by  the  sign  of  the  coordinate. 
Through  the  points  thus  determined  draw  lines  parallel 


Fig.  8. 


10 


ANALYTIC   GEOMETRY 


[Ch.  II,  §  0 


to  the  axes,  and  their  intersection  will  locate  the  point 
whose  coordinates  were  given.  The  following  figures 
illustrate  the  method.  Coordinate  paper  having  two  per- 
pendicular sets  of  parallel  lines  is  very  useful,  and  should 
be  obtained  by  the  student. 


C-A,  7) 


0 


(-2,-1 


XS,8) 


<x 


(7,-5) 


Y 
Fig.  9. 


Fig.  10. 


PROBLEMS 

1.  Plot  the  following  points : 

(0,  0),  (0,  -3),  (4,  0),  (-4,  0),  (-4,  6),  (-3,  -8). 

2.  Construct  the"  quadrilateral  whose  vertices  are  the  points 
(7,  2),  (0,  -  9),  (-  3,  -  1),  and  (-  6,  4). 

3.  What  relation  exists  between  the  coordinates  of  two 
points  if  the  line  joining  them  is  bisected  at  the  origin  ? 

4.  What  are  the  coordinates  of  the  corners  of  a  square 
whose  side  is  s,  if  the  origin  is  at  the  centre  of  the  square  and 
(a)  the  axes  are  parallel  to  the  sides,  (b)  the  axes  coincide 
with  the  diagonals  ? 

5.  If  one  side  of  a  parallelogram  coincides  with  the  X-axis 
and  one  vertex  is  at  the  origin,  express  in  the  simplest  way 
the  coordinates  of  the  other  vertices,  (a)  in  rectangular  coor- 
dinates, (b)  in  oblique  coordinates. 

6.  What  are  the  coordinates  of  the  vertices  of  an  equi- 
lateral triangle,  if  (a)  one  side  coincides  with  the  X-axis  and 


Ch.  II,  §§  10,  11] 


THE    POINT 


11 


the  origin  is  at  one  vertex,  (6)  one  side  coincides  with  the 
.X-axis  and  the  origin  is  at  the  middle  of  this  side,  (c)  if  the 
origin  is  at  the  centre  of  the  triangle  and  the  X-axis  passes 
through  one  vertex  ? 

10.  Notation.  —  It  will  often  be  necessary  to  distinguish 
between  points  which  are  fixed  and  those  which,  although 
constrained  to  move  in  a  certain  path,  yet  can  occupy 
any  position  along  this  path.  Fixed  points  will  always 
be  distinguished  by  means  of  subscripts,  being  lettered 
Pv  PT  etc.,  and  represented  by  the  coordinates  (xv  y^), 
(.r2,  t/2),  etc.,  while  variable  points  will  generally  be  rep- 
resented by  the  simple  variables  (x,  y).  If  variable 
points  whose  movements  are  governed  by  different  laws 
are  under  consideration  at  the  same  time,  they  will  be 
distinguished  by  using  (x,  ?/),  (V,  ?/'),  etc. 

11.  Distance  between  two  points  in  rectangular  coordi- 
nates. —  One  of  the  first  questions  which  naturally  arises 
concerning  points  is  that  of  finding  the  distance  between 
them  when  their  coordinates  are  given. 

Let  Px  and  P2  (in  either  figure)  be  two  points  wdiose 
coordinates  are  (xv  y{)  and  (x2,  y2).     Drop  perpendicu- 


M. 


Y 
Fig.  11. 


M, 


-A' 


YZ  ANALYTIC   GEOMETRY  [Ch.  II,  §  11 

i 
lars  on  the  X-axis  and  draw  P2K  to  meet  MlPl  or  that 

line  produced.  Then  xl  =  0MV  x2  =  0M2,  yx  =  MXPV 
and  y2  =  M2P2.  It  must  be  remembered  that  the  coordi- 
nates of  any  point  are  measured  from  the  coordinate  axes 
and  must  be  so  read. 

In  either  figure  PXP2  =  V/yf*  +  KP?. 
But      P2K=  M2MX  =  0M1  -  0M2  =  x1-  xv 
and  KP1  =  MlPl  -  i%K=  3I.P,  -  M2P2  =  yx  -  y2. 

Hence  PXP,  =  V(asi  -  oc2)2  4-  (*/i  -  2/2/2.  [1] 

This  being  a  length  merely,  it  is  immaterial  whether 
it  is  read  PXP2  or  P2PV 

It  is  necessary  for  the  student  to  make  himself  familiar 
at  once  with  demonstrations  of  this  kind  in  which  a  single 
demonstration  will  ajiply  to  all  possible  cases.  It  might  seem 
at  first  that  in  Fig.  12  the  equation  KPt =M1P1  —  MXK 
does  not  hold.  But  if  MXK  is  replaced  by  its  equal 
—  KMV  the  equation  is  at  once  seen  to  be  true. 

Let  the  student  draw  various  figures  with  the  points 
in  different  quadrants,  and  assure  himself  that  the  same 
demonstration  holds  for  all.  Care  must  be  taken  to  read 
the  lines  always  in  the  proper  direction.  For  simplicity 
the  figures  will  usually  be  constructed  in  the  first  quad- 
rant, but  the  student  should  always  satisfy  himself  that 
there  is  no  restriction  on  their  position,  and  that,  if  any 
other  figure  is  constructed  and  lettered  in  a  correspond- 
ing way,  just  the  same  demonstration  will  hold  letter 
for  letter. 


Oh.  II,  §  12] 


THE    POINT 


13 


12.  Distance  between  two  points  in  oblique  coordinates.  — 
When  the  axes  are  oblique,  draw  M1P1  and  M2P2,  the 
ordinates  of  Px  and  jP2,  and  the  line  P2K  parallel  to 
the  X-axis.  Since  P2K  and  KPX  are  to  be  expressed 
in  terms  of  the  coordinates  of  Px  and  P2,  their  positive 


Fig.  13. 


Fig.  14. 


directions  will  be  the  same  as  the  positive  directions  of 
the  axes.  The  angle  between  them  will  always  be  a>, 
and  the  generalized  form  of  the  law  of  the  cosines 
(Art.  7)  gives 


PXP2  =  Vp2jf2  +  jzpf  +  2  P2K-  KPX  cos  *», 

where  not  only  the  magnitudes,  but  also  the  directions 
of  the  lines  are  considered.     But 

P2K=  M2MX  =  0M1  -  0M2  =  xx-  x2, 

and    KP^M^-M^M^-MzP^y^yv 

Substituting  these  values,  we  have 
P\P%  =  V(a?!  -  a?2)2  +  (2/i  -  Z/2)'2  +  2(«i  -  a?2)(2/i  -  y-i)  cos  «,     [2] 
as  the  distance  between  two  points  in  oblique  coordinates. 


14  ANALYTIC   GEOMETRY  [Ch.  II,  §  13 

PROBLEMS 

1.  Find  the  distance  between  the  two  points  whose  rec- 
tangular coordinates  are  (—  2,  6)  and  (1,  5). 

Solution.  — In  using  formulas  [1]  and  [2]  we  may  choose  either  of 
the  points  for  Pi  and  the  other  for  P2.  Let  (-2,  6)  he  the  coordinates 
of  Pi,  and  (1,  5)  the  coordinates  of  P2. 

Then  Pi P2  =  V(- 2  -  1)* +(tt  -  5)*  =  VlO. 

2.  Find  the  lengths  of  the  sides  of  a  triangle  if  the  rec- 
tangular coordinates  of  its  vertices  are  (—  3,  4),  (—  G,—  1),  and 

(4,-5). 

3.  Find  the  lengths  of  the  sides  of  the  triangle,  the  coordi- 
nates of  whose  vertices,  referred  to  axes  making  an  angle  of 
60°  with  each  other,  are  (0,  0),  (-  5,  -  5),  and  (1,  -  3). 

4.  What  is  the  distance  from  the  origin  to  the  point  (a,  b) 
in  rectangular  coordinates  ?  In  oblique  coordinates,  if  the 
angle  between  the  axes  is  45°  ? 

5.  Show  that  the  points  (G,  4),  (2,  8),  (3,  -  2),  and  (-  1,  2) 
are  the  vertices  of  a  parallelogram. 

6.  Show  that  the  lengths  of  the  diagonals  of  any  rectangle 
are  equal. 

Note.  —  Take  the  two  adjacent  sides  as  axes  and  call  the  opposite 
vertex  («,  b). 

13.  Points  dividing  a  line  in  a  given  ratio. — The  next 
question  to  be  discussed  is  that  of  finding  the  coordinates 
of  a  point  which  will  divide  the  line  joining  two  given 
points  in  any  given  ratio.  We  must  first  define  what  we 
mean  by  "dividing  the  line  joining  two  points  in  any 
given  ratio  " ;  for  it  has  a  larger  meaning  here  than  we 
have  been  accustomed  to  give  it. 

If  C  is  any  point  on  the  line  AB,  it  is  said  to  divide  the 


Cm.  II,  J  13] 


THE    TOIXT 


15 


lint'  AB  into  the  two  parts  AC  and  CB  (care  being  taken 

to  read  the  two  parts  in  just  this  way)  whether  the  point 

C  lies  between  A  and  B  or 

beyond   either.       It  will  be     — 

seen  that  if  the  point  0  lies 

between  A  and  £,  the  ratio,  77^,  of  the  parts  into  which 

it  divides  the  line  is  positive  ;  while  if  it  lies  on  AB  pro 
ducecl,  the  ratio  is  negative 


Fig.  15. 


AC 

If  has  a  value  between 


CB 

AC 
1  and  —  1,  C  is  nearer  A,  while  if  ■-—   is  greater  than  1 

or  less  than  —  1,  C  is  nearer  B. 

We  shall  now  obtain  the  formulas  for  finding  the  coor- 
dinates of  the  point  P  which  divides  the  line  PXP2  *n  ^ne 


ratio  m1 :  m<A 


,,    ,    P,P      m, 
or  so  that  -— \- -  =  - 
PI 


2 

in  2 

F 

r 

7T 

K,1- 

s£ 

,\ 

!      >y 

0 

V 

XM 

Mf        \ 

v' 

Fig.  16. 


Fig.  17. 


Draw  the  ordinates  MtPv  M2P2,  and  MP.  Also  draw 
the  lines  PXKX  and  PiT  parallel  to  the  X-axis.  The 
triangles  PPXKX  and  PKP2  are  evidently  similar,  and 


P^  =  K,P_P,P 
PK      KPo     PP. 


rn  ■ 


16  ANALYTIC   GEOMETRY  [Ch.  II,  §  14 

But  PXKX  =  0M    -  0MX  =  x  -  xv 

PK    =  0M2    -0M   =x2-z, 
KXP   =MP     -MK1=y  ~yv 
KP2  =  M2P2  -  M2K=y2  -  y. 

Substituting  these  values,  we  have 

x-x1=dm^  and  y-y1  =  wi 
x2-  x     ma  y2  —  y     m2 

Solving,    »  =  ""»  + *"*,    and    ,=—*  +  »■*.        [3] 

If  the  point  P  bisects  the  line  P±PV  m1  —  ra2,  and  the 
formulas  become 

«  =  &±S,  and  „  =  Ki±ife.  [4] 

Let  the  student  go  over  the  demonstration  carefully, 
using  the  second  figure,  and  assure  himself  that  every 
step  holds  as  well  for  that  as  for  the  first.  Let  him  also 
construct  other  figures  with  the  points  in  different  posi- 
tions, but  using  the  same  letters  for  corresponding  points. 

Since  the  demonstration  depends  only  upon  the  simi- 
larity of  triangles,  it  will  hold  also  in  oblique  coordinates. 
The  results  are  therefore  general,  and  will  apply  to  either 
system  of  Cartesian  coordinates. 

14.    Harmonic  division.  —  If  the  line  A  C  is  divided  by 

the  points  B  and  7),  internally 

— ' ' .     and    externally,     in    the    same 

Fig.  18.  •      ,  , .  ,, 

numerical     ratio,     or    so     that 

— -  =  —  — — ,  the  line   .AC  is   said   to   be   divided    har- 
Jo  O  AJ  o 

monically. 


Ch.  II,  §  14] 


THE   POINT 


17 


Let  the  student  show  that  the  line   BD  will  then  be 
divided  harmonically  by  the  points  C  and  A,  or  so  that 

BC=     BA 
CD         AD 

The   four  points  A,  B,    (7,  and  D  are  said  to  form  a 
harmonic  range. 

If  parallel  lines  are  drawn  through  the  points  A,  B,  C, 
and    D    of    a    harmonic    range, 
their    intersections    A',    B' ,    C\ 
and    D',    with    any    transversal, 
will  also  be  a  harmonic   range. 

For,    from     plane     geometry, 


AB 
BO 


A'B' 

B'C 

Hence 


and 


AD     A'D' 


A 

B 

C 

A^- — * 

B^. 

9^-^ 

DC     D'C 
A'B' 


B'C 


A'D' 
D'C 


Fig.  19. 


D 


PROBLEMS 


1.    Find    the    points    of     trisection    of    the    line     joining 
(_  3,  _  4)  and  (5,  2). 

Solution.  —  If  we  wish  to  find  P,  the  point  of  trisection  nearest  P>, 


m\X2  +  m^xi 

mjy2  +  nioyx 
mi  +  wi2 


2 

5  +  l(- 

-3) 

1+2 

2 

2  +  l(- 

-4) 

<). 


1  +  2 

and  the  point  of  trisection  is  (|,  0). 

But  if  we  wish  to  find  P',  mi  =  1,  m®  =  2, 

_  1 
3' 


a,  =  1.5  +  2(-3) 


1.2  +  2(-4) 
y  1  +  2 

and  the  other  point  of  trisection  is  (—  |,  —  2). 


18  ANALYTIC   GEOMETRY  [Ch.  II,  §  14 

2.  Find  the  point  which  divides  the  line  through  (—3,  —4) 
and  (5,  2)  in  the  ratio  —  }. 

3.  Extend  the  line  through  (1,  5)  and  (—3,  4)  beyond  the 
latter  point  until  it  is  three  times  its  original  length.  Find 
the  coordinates  of  its  extremity. 

4.  In  the  triangle  whose  vertices  are  (0,  0),  (0,  6),  (5,  8), 
find  the  point  on  each  median  which  is  two-thirds  of  the  dis- 
tance from  the  vertex  to  the  middle  point  of  the  opposite  side, 
and  show  that  these  points  coincide. 

5.  Show  that  the  medians  of  any  triangle  meet  in  a  point, 
choosing  the  axes  so  that  the  vertices  may  be  represented  by 
(0,  0),  (a,  0),  and  (b,  c). 

6.  In  the  right  triangle  whose  vertices  are  (0, 0),  (0,  6),  and 
(8,  0),  show  that  the  distance  from  the  vertex  of  the  right 
angle  to  the  middle  point  of  the  opposite  side  is  equal  to  one- 
half  of  the  hypotenuse. 

7.  Prove  that  the  theorem  of  problem  6  holds  for  any  right 
triangle. 

Note.  — Take  the  legs  of  the  triangle  as  axes. 

8.  In  the  triangle  whose  vertices  are  A  (—  1,  2),  B  (4,  5), 
and  (7(3,  —.4),  a  line  DE  is  drawn  through  the  middle  points 
of  the  sides  AB  and  AC.     Show  that  BC  =  2  DE. 

9.  Prove  that  the  line  joining  the  middle  points  of  the  sides 
of  a  triangle  is  equal  to  one-half  of  the  third  side,  using  the 
points  (xlf  2/j),  (x2  y2),  and  (x3,  ys)  as  the  vertices  of  the  triangle. 

10.  If  the  coordinates  of  three  oMhe  vertices  of  a  parallelo- 
gram are  (0,  0),  (8,  0),  and  (3,  5),  find  the  coordinates  of  the 
fourth  vertex,  which  lies  in  the  first  quadrant. 

11.  Prove  that  the  diagonals  of  any  parallelogram  bisect 
each  other. 

12.  In  what  ratio  is  the  line  joining  the  points  (—  1,  6)  and 
(7,  -  2)  divided  by  the  point  (2,  3)  ?  by  the  point  (10,  -  5)? 


Ch.  II,  §  14]  THE   POINT  19 

13.  The  line  joining  the  points  (0,  3)  and  (9,  0)  is  divided 
internally  by  the  point  (3,  2).  Find  the  coordinates  of  the 
point  which  divides  it  externally  in  the  same  numerical  ratio. 

14.  Find  the  coordinates  of  the  point  P  which  forms,  with 
the  points  A  (4, 1),  B  (2,  -  2),  and  C  (-  2,  -  8),  a  harmonic 
range,  if  (a)  P  is  between  A  and  72;   (b)  P  is  between  B  and  C. 


CHAPTER   III 
LOCI 

15.  Equation  of  a  locus.  —  In  the  previous  chapter  we 
have  considered  fixed  points  only.  If  a  point  is  made  to 
move  in  the  plane  according  to  some  definite  law,  a  curve 
or  locus  is  generated.  (The  term  "  curve  "  in  Analytic 
Geometry  is  applied  to  any  locus,  including  straight 
lines.)  As,  for  example,  a  point  which  remains  at  a  fixed 
distance  from  a  given  fixed  point  generates  a  locus  called 
a  circle  ;  a  point  which  is  always  equally  distant  from 
two  intersecting  lines  generates  a  locus  which  is  the 
bisector  of  the  angle  between  those  lines ;  a  point  which 
is  always  equally  distant  from  the  ends  of  a  line  generates 
the  perpendicular  bisector  of  that  line,  etc. 

If  now  we  can  translate  the  statement  of  the  law  govern- 
ing the  movement  of  a  point  into  an  algebraic  relation  or 
equation  between  the  coordinates  of  the  points  which  satisfy 
the  law,  we  shall  have  an  equation  which  may  be  used  to 
represent  the  curve.  For,  if  our  translation  is  correct, 
every  point  whose  coordinates  satisfy  the  equation  will 
occupy  a  position  on  the  path  generated  by  the  moving 
point,  since  the  equation  is  only  a  restatement  of  the  law 
itself  in  algebraic  language.  There  will  be,  moreover,  no 
position  of  the  moving  point  whose  coordinates  do  not 
satisfy  the  equation.  We  shall  then  have  obtained  an 
equation  which  is   satisfied  by  the  coordinates  of  every 

20 


Jf 


Ch.  Ill,  §  15]  LOCI  21 

point  on  the  locus,  and  by  the  coordinates  of   no  point 
not  on  the  locus. 

In  the  first  example  given  above,  if  the  fixed  point  is 
taken  as  the  origin,  and  if  the  moving  point  P  remains 
at  a  distance  a  from  the  fixed  point,  the  relation  between 
the  coordinates  x  and  y  of  every  position  of  P  is  x2  +  y2=  a2. 
For  (Fig.  21)  Oil2  +  MP2  =  OP2. 

We  see,  moreover,  that  this 
equation  cannot  be  satisfied  by 
any  point  which  is  not  at  a  dis- 
tance a  from  0.     We  have  then 

X- 

translated  the    given  condition 

into   algebraic   language.      The 

equation  and  the  curve  bear  to 

each  other  the  following  recip-  [y' 

rocal  relation  :    The  coordinates  Fig.  21. 

of  every  point  on  the  circle  satisfy 

the  equation,  and  conversely,  every  point  whose  coordinates 

satisfy  the  equation  lies  on  the  circle.      When  an  equation 

and  a  curve  are  connected  by  this  relation,  the  equation 

is  spoken  of  as  the  equation  of  the  curve,  and  the  curve  as 

the  locus  of  the  equation. 

Again,  let  a  point  move  so  as  to  remain  equally  distant 
from  the  two  axes.  What  is  the  algebraic  translation  of 
this  law,  or,  in  other  words,  what  is  the  algebraic  equation 
which  must  be  satisfied  by  the  coordinates  of  every  point 
governed  by  the  law  ?  It  is  evidently  x  =  y,  and  this  is, 
therefore,  the  equation  of  the  bisector  of  the  angle  be- 
tween OX  and  OY. 

What  is  the  equation  of  the  bisector  of  the  angle 
between  OY  and  OX' 1 


oo 


ANALYTIC    GEOMETRY 


[Ch.  Ill,  §  15 


If  a  point  moves  so  as  to  be  always  three  units  above 
the  X-axis,  the  ordinate  of  every  point  must  be  three, 
while  no  restriction  is  placed  on  the  abscissa  of  the  point. 
This  law,  translated  into  algebraic  language,  is,  therefore, 
y  =  3 ;  for  this  equation  makes  just  the  same  statement 
in  regard  to  the  position  of  every  point  which  satisfies  it. 
What  is  the  equation  of  the  locus  of  points  two  units 
to  the  left  of  the  Y"-axis  ? 

What  are  the  equations  of  the  axes  ? 
The  third  illustration  was  the  locus  of  a  point  which 
moves  so  as  always  to  be  equally  distant  from  two  fixed 
points. 

Place  the  axes  with  the  origin  at  one  of  the  points,  and 
the  X-axis  coincident  with  the  line  joining  the  two  points. 

Let  .the  distance  OA  (Fig.  22) 
between  the  two  points  be  rep- 
resented by  a.  Then  the  co5r- 
dinates^  of  the  two  points  are 
(0,  0)  and  (a,  0).  We  can 
translate  into  an  algebraic  equa- 
tion the  statement  that  a  point 
P,  coordinates  (#,  ?/),  shall  be 
equally  distant  from  the  two 
fixed  points  0  and  J.,  by  ex- 
pressing the  distances  of  P  from  each  of  the  two  points, 
and  equating  these  two  expressions. 


In  Fig.  22, 

OP  =  V*2  +  y\ 

By  [i] 

and 

AP  =  VO  -  a)2  +  f. 

Bj[l] 

Equating, 

Vz2  +  */2  =  V<>-«)2  +  ?/2. 

Ch.  Ill,  §  15]  Loci  23 

Squaring  and  reducing,  we  have,  as  the  equation  of  the 

desired  locus, 

a 

x  =  ~i 

Here  the  final  result  does  not  express  so  clearly  as  in  the 
previous  cases  that  it  is  simply  a  translation  of  the  state- 
ment of  the  law.  But  it  has  been  obtained  by  simple 
algebraic  reductions  from  this  exact  statement.  The 
result  is,  as  we  should  expect,  the  perpendicular  bisector 
of  the  line  joining  the  two  fixed  points. 

We  have  in  these  simple  cases  been  able  to  translate 
the  law  governing  the  movement  of  a  point  in  the  plane 
into  an  algebraic  equation.  There  are  many  loci  for 
which  this  is  possible.  But  the  law  may  be  stated  in 
such  a  way  as  to  require  other  than  algebraic  symbols  to 
represent  it.  For  example,  the  path  of  any  fixed  point 
on  the  circumference  of  a  wheel  rolling  on  a  straight  line 
in  a  plane  is  a  perfectly  definite  curve.  But  the  relation 
between  the  coordinates  cannot  be  expressed  in  a  single 
algebraic  equation.  It  requires  the  introduction  of  trigo- 
nometric functions. 

If  a  point  moves  at  random,  no  equation  connecting  the 
coordinates  of  its  different  positions  can  be  found ;  for  an 
equation  imposes  a  law  upon  the  movement  of  the  point. 

PROBLEMS 

1.  Find  the  equation  of  the  locus  of  points  which  are 
equally  distant  from  the  points  (1,  3)  and  (—2,  5). 

2.  Find  the  equation  of  the  locus  of  points  which  are  three 
times  as  far  from  the  X-axis  as  from  the  l^axis. 

3.  Find  the  equation  of  the  locus  of  points  which  are  five 
units  from  the  point  (—3,  4). 


24  ANALYTIC   GEOMETRY  [Ch.  Ill,  §  16 

4.  A  point  moves  so  as  to  be  always  five  times  as  far  from 
the  r"-axis  as  from  the  point  (5,  0).  Find  the  equation  of  its 
locus. 

5.  A  point  moves  so  that  the  sum  of  the  squares  of  its 
distances  from  the  points  (0,  0)  and  (5,  —  5)  is  always  equal  to 
40.     Find  the  equation  of  its  locus. 

6.  A  point  moves  so  as  to  be  always  three  times  as  far 
from  the  point  (1,  —  2)  as  from  the  point  (—3,  4).  Find  the 
equation  of  its  locus. 

7.  A  point  moves  so  that  the  sum  of  its  distances  from  the 
two  axes  is  always  equal  to  10.     Find  the  equation  of  its  locus. 

8.  A  point  moves  so  that  its  distance  from  the  X-axis  is 
always  one-half  its  distance  from  the  origin.  Find  the  equation 
of  its  locus. 

9.  A  point  moves  so  that  its  distance  from  the  point 
(—4,  1)  is  always  equal  to  its  distance  from  the  origin.  Find 
the  equation  of  its  locus. 

10.  A  point  moves  so  that  the  square  of  its  distance  from 
the  origin  is  always  equal  to  the  sum  of  its  distances  from  the 
axes.     Find  the  equation  of  its  locus. 

16.  Locus  of  an  equation.  —  Looking  at  the  question 
from  the  other  side,  let  us  consider  what  will  be  the 
geometric  interpretation  of  any  given  equation  in  x  and  y. 
It  is  at  once  evident  that  only  the  coordinates  of  certain 
points  in  the  plane  will  satisfy  the  equation ;  for,  if  we 
give  any  particular  value  to  x,  one  or  more  values  of  y 
will  be  determined.  The  point,  then,  cannot  occupy  any 
position  at  random  in  the  plane,  yet  it  is  not  confined  to 
a  finite  number  of  positions.  For,  since  any  value  we 
please  may  be  assigned  to  x,  there  will  be  an  indefinite 
number  of   positions  whose  coordinates  will  satisfy  the 


Ch.  Ill,  §  17]  LOCI  25 

equation.  Moreover,  it  appears  that  these  points  are  not 
scattered  indiscriminately  over  the  plane,  since  random 
values  of  x  and  y  will  not  satisfy  the  equation.  Small 
changes  in  the  value  of  x  will  in  general  produce  small 
changes  in  the  value  of  y.  Points  may  therefore  be 
found  as  close  as  we  please  to  each  other,  and  from  this 
we  may  infer  that  they  are  situated  on  some  curve.  This 
curve  which  contains  all  the  points  which  satisfy  the  equation 
and  no  others  is  called  the  locus  of  the  equation. 

17.  Plotting  the  locus  of  an  equation.  —  How  shall  we 
determine  the  locus  of  any  given  equation  ?  Sometimes 
the  locus  is  at  once  evident.  For  example,  what  is  the 
geometric  interpretation  of  the  equation  y  =  3  ?  The 
equation  says  nothing  concerning  the  abscissas  of  points 
on  the  locus,  but  fixes  the  ordinate  of  every  point.  All 
points  which  satisfy  it  must  therefore  lie  at  a  distance  of 
three  units  above  the  X-axis.  Hence  the  locus  is  a  line 
parallel  to  the  X-axis,  and  three  units  above  it. 

Again,  consider  the  equation  x  —  y.  It  states  in  alge- 
braic language  that  a  point  moves  so  as  to  remain  equally 
distant  from  the  two  axes.  Its  locus  is  therefore  the  line 
which  bisects  the  angle  between  the  two  axes. 

Sometimes  it  is  easy,  as  in  these  cases,  to  translate  the 
algebraic  equation  into  the  law  which  governs  the  move- 
ment of  the  point,  and  hence  determine  the  exact  form 
and  position  of  the  locus.  But  this  is  often  difficult,  and 
we  must  have  other  means  of  determining  the  curve.  We 
can  always  determine  as  many  points  as  we  please  on  the 
locus  by  giving  to  one  of  the  coordinates  a  series  of  values 
and  determining  the  corresponding  values  of  the  other. 


26 


ANALYTIC   GEOMETRY 


[Ch.  Ill,  §  17 


Place  these  points  in  their  proper  positions  in  the  plane, 
and  when  a  sufficient  number  has  been  obtained,  a  smooth 
curve  passed  through  them  will  show  approximately  the 
form  of  the  curve.  The  points  can  be  determined  as  near 
to  each  other  as  we  please,  and  the  approximation  can  be 
carried  to  any  required  degree  of  accuracy.  This  is  called 
plotting  the  curve. 

We  shall  plot  the  locus  of  the  equation 

Give  consecutive  values  to  x,  and  find  the  corresponding 
values  of  y. 

If 


x  =  0, 

y  =  10, 

x=l, 

y=  8, 

x=% 

9f=   6, 

x=8, 

y=  4, 

z  =  4, 

y  =  2, 

x  =  5, 

y=  o, 

x  =  - 

i, 

y  =  i2, 

x  =  - 

2, 

y  =  14, 

x—  — 

3, 

2/ =  16, 

etc. 

etc. 

Plotting  the  points  (0,  10),  (1,  8),  (2,  6),  etc.,  and 
passing  a  curve  through  the  points,  we  see  that  they  all 
appear  to  lie  on  a  straight  line.  This  method,  however, 
does  not  assure  us  that  the  locus  is  a  straight  line.  It 
only  shows  that,  so  far  as  our  construction  is  accurate,  it 
appears  to  be  a  straight  line. 

We  shall  show  later  that  every  equation  of  the  first 
degree  represents  a  straight  line. 


Ch.  Ill,  §  18] 


LOCI 


27 


Again,  let  us  plot  the  locus  of  the  equation 

25. 


x*-y* 


Solving  the  equation  for  y,  we  have  y  =  ±  Va?2  —  25, 
from  which  it  appears  that  y  is  imaginary,  so  long  as 
—  5  <  x  <  +  5.  There  will  therefore  be  no  points  on  the 
locus  for  which  x  is  numerically  less  than  5. 

If     x  —  5,  y  —  0 1  x  =  —  5,  y  —  0 ; 

x  =  6,   y  =  ±VIlj  x  =  -6,    ?/  =  ±Vll; 


#  =  7,   y  ==  ±  V24 


etc. 


Plotting  the  points  (5,  0),  ((3,  +  Vll),  (6,  -  Vll),  etc., 
and    passing    a    smooth 
curve  through  them,  we  y 

have  the  curve  in  Fig. 
24.  It  can  be  seen  from 
the  equation  that  each 
branch  goes  off  indefi- 
nitely, never  again  turn- 
ing toward  either  axis ; 
for  as  x  increases,  y  in- 
creases indefinitely. 

18.  Symmetry.  —  A  curve  is  said  to  be  symmetrical 
with  respect  to  one  of  two  axes  (rectangular  or  oblique) 
when  that  axis  bisects  every  chord  parallel  to  the  other. 

A  curve  is  said  to  be  symmetrical  with  respect  to 
a  point  when  that  point  bisects  every  chord  drawn 
through    it. 

It  is  easily  proved  that  if  a  curve  is  symmetrical  with 
respect   to  two  axes,   it  is  symmetrical  with  respect  to 


28  ANALYTIC   GEOMETRY  [Ch.  Ill,  §  18 

their  point  of  intersection.  Now,  if,  upon  substituting 
any  value  for  x  in  an  equation,  we  find  two  values  of  ?/, 
equal  numerically  but  with  opposite  signs,  the  curve  is 
evidently  symmetrical  with  respect  to  the  X-axis.  Or, 
if,  for  every  value  of  ?/,  we  find  two  values  of  x,  equal 
numerically  but  with  opposite  signs,  the  curve  is  evi- 
dently symmetrical  with  respect  to  the  T^-axis.  If  both 
these  occur,  the  curve  must  be  symmetrical  with  respect 
to  the  origin. 

It  appears  that  the  first  of  these  conditions  can  be 
satisfied  when  y  occurs  in  the  equation  in  even  powers 
only,  and  the  second  when  x  occurs  in  even  powers  only. 
A  curve  is  therefore  symmetrical  with  respect  to  the  X-axis 
tvhen  its  equation  does  not  contain  odd  powers  of  y ;  it  is 
symmetrical  with  respect  to  the  Y-axis  ivhen  its  equation 
does  not  contain  odd  powers  of  x. 

It  is  symmetrical  with  respect  to  the  origin  if  its  equation 
contains  no  term  of  an  odd  degree  in  x  and  y. 

We  can  therefore  tell  at  once  whether  a  curve  is  sym- 
metrical with  respect  to  either  or  both  axes.  This  is 
useful  in  plotting ;  for  if  a  curve  is  symmetrical  with 
respect  to  the  Jf-axis,  it  is  only  necessary  to  plot  the  part 
above  that  axis  and  form  the  same  curve  below ;  if  sym- 
metrical with  respect  to  the  I^-axis,  to  plot  the  part  at 
the  right  of  that  axis  and  form  the  same  curve  at  the 
left. 

The  curve  which  we  have  just  plotted,  x2  —  y2  =  25, 
is  evidently  symmetrical  with  respect  to  both  axes.  It 
would  have  been  sufficient  to  have  plotted  that  part  which 
lies  in  the  first  quadrant  and  determined  the  rest  of  the 
curve  from  this, 


Ch.  Ill,  §  18]  LOCI  29 

PROBLEMS 

1.  Plot  the  loci  of  the  following  equations : 

(a)  x2  +  y2  =  9-  (/)  4  a?  +  9  ?/2  =  0. 

(b)  x2-\-7f  =  0.  (g)  4or-9/  =  0. 
(C)  ^-^  =  0.                       (/t)  i^  =  4a;. 

(d)  4  a2  +  9  f  =  36.  (i)    a2  =  4  y. 

(e)  4ar,-9</2  =  36.  (j)  y2  =  -±x. 

2.  Plot  the  locus  of  the  equation 

a2  +  2/-4a;  +  4y-12=0. 

The  form  of  the  equation  shows  at  once  that  the  curve  is 
not  symmetrical  with  respect  to  either  axis.  Solving  for  x  in 
terms  of  y,  we  have 

x2-±x  =  l2-±y-2y2, 


or  x  =  2±V16-4:y-2y2. 

From  which  it  appears  that  the  locus  is  symmetrical  with 
respect  to  the  line  x  =  2. 

There  will  be  real  values  of  x  only  for  those  values  of  y  which 
make  16  —  4  y  —  2  y2  positive  or  zero.  This  expression  vanishes 
when  y  =  2,  or  —  4,  and  can  be  factored  into  (4  +  y)  (4  —  2  y). 
It  is  evidently  positive  when  —  4  <  y  <  2,  and  negative  for  all 
other  values  of  y. 

Solving  the  given  equation  for  y  in  terms  of  x,  we  have 


jU  +  lx-x2 

From  which  it  appears  that  the  locus  is  symmetrical  with 
respect  to  the  line  y  =  —  1. 

The  expression  14-+-4^  —  x2  vanishes  when  x  =  2  ±3V2, 
and  can  be  factored  into  (2  +  3  V2  —  x)  (—  2  +  3  V2  +  x).  It  is 
evidently  positive  when  2  —  3  V2  <  x  <  2  +  3  V2,  and  negative 
for  all  other  values  a;.  There  are  then  real  points  on  the  locus 
only  when  2  -  3  V2  <  x  <  2  +  3  V2,  and  -  4  <  y  <  2. 


30 


ANALYTIC   GEOMETRY 


[Ch.  Ill,  §  18 


The  curve  is  therefore  symmetrical  with  respect  to  the  two 
lines  x  =  2  and  y  =  —  1,  and  lies  wholly  within  the  four  lines 
<b=2-3V2,  z  =  2+3V2,  y  =  -±,  and  y  =  2. 

Giving  y  the  values  —  4,  —  3,  —  2,  —  1,  0,  1,  and  2,  we  have 
the  following  points  on  the  locus : 

(2,  -4),   (2±ViO,  -3),   (G,  -2),   (-2,  -2), 

(2  ±  3  V2,  -  1),   (6,  0),   (-  2,  0),   (2  ±  VlO,  1),    (2,  2). 

Plotting  these  points  and  drawing  a  smooth  curve  through 
them,  we  have  a  fairly  clear  notion  of  the  form  of  the  locus. 


x 


0 

X 


T 

Fig.  25. 


3.    Plot  the  loci  of  the  following  equations : 


(a)  z2  +  2/2  +  2a--2?/-10  =  0. 

(b)  x2  +  y2-x-8  =  0. 

(c)  a^  +  ^-4y  +  15  =  0. 

(d)  4z2  +  9?/2-8?/-6=0. 

(e)  a?y  =  0. 
(/)  ay  =100. 


(g)   x-  =  y\ 
(k)  y  =  x\ 
(i)    y  =  sin  x. 
0")  2/  =  tan  x. 
(Jc)  y  =  sec  x.* 
(Z)  y  =  cos-1  2-. 


Cn.  Ill,  §§  10-21] 


LOCI 


31 


19.  Intercepts.  —  The  distances  from  the  origin  to  the 
points  where  a  curve  cuts  the  axes  are  called  the  intercepts 
of  the  curve. 

One  of  the  coordinates  of  such  a  point  will  always  be 
zero  and  the  other  will  be  the  intercept.  Hence  the  in- 
tercepts on  the  X-axis  can  be  found  by  substituting 
y  =  0  in  the  equation  and  finding  the  corresponding  values 
of  x ;  the  intercepts  on  the  y-axis,  by  substituting  x  =  0 
and  finding  the  corresponding  values  of  y. 

20.  Intersection  of  two  curves. — When  two  curves  in- 
tersect,  the  coordinates  of  the  point  of  intersection  must 
satisfy  both  equations.  In  order  to  find  the  coordinates 
of  such  a  point  of  intersection,  it  is  only  necessary  to  find 
the  values  of  x  and  y  which  will  satisfy  both  equations,  or 
in  other  words,  to  solve  the  equations  simultaneously. 


21.  Locus  of  u  +  kv  =o  and  uv  =  o.  — If  all  the  terms 
of  an  equation  are  transposed  to  the  first  member,  we  may 
represent  them  by  a  single  letter,  as  u  or  t>,  and  speak  of 
the  equation  as  ?/  =  0  or 
v  =  0.  The  letters  u  and 
v  are  simply  used  as  ab- 
breviations for  expres- 
sions in  x  and  y  of  any 
degree.  Then  u  =  0  will 
represent  some  curve,  and 
v  =  0  another  curve.  Let 
us  consider  what  will 
be  represented  by  the 
equation  u  -h  kv  =  0,  where  Fig.  26. 


32  ANALYTIC    GEOMETRY  [Ch.  Ill,  §  21 

k  is  any  constant  quantity,  positive  or  negative.  Let 
(xv  yx)  be  any  point  of  intersection  of  the  two  curves 
u  —  0  and  v  =  0.  Its  coordinates  will  satisfy  both  these 
equations,  and  hence  will  satisfy  the  equation  u  +  kv  =  0. 
The  locus  of  u  -f  kv  =  0  must  therefore  pass  through  all  the 
points  common  to  the  two  curves  u  =  0  and  v  =  0.  More- 
over, it  will  not  pass  through  any  other  point  of  either 
curve.  For  the  coordinates  of  any  such  point  will  cause 
one  of  the  expressions  u  or  v  to  vanish,  but  not  the  other, 
and  therefore  cannot  satisfy  the  equation  u  -f-  kv  =  0. 

Again,  let  us  consider  what  will  be  represented  by  the 
equation  uv  =  0.  It  is  evident  that  the  coordinates  of 
every  point  which  cause  either  u  or  v  to  vanish  will  satisfy 
this  equation,  and  that  the  coordinates  of  no  other  point 
can  satisfy  it.  uv  =  0  must  therefore  represent  the  loci  of 
the  two  equations  u  =  0  and  v  =  0,  taken  together.  For 
example,  xy  —  0  represents  both  coordinate  axes. 

PROBLEMS 

1.  Find  the  intercepts  of  the  curves  whose  equations  are 
given  on  page  29. 

2.  Find  the  points  of  intersection  of  the  following  curves : 

(a)    x2  -f-  if  =  25  and  x  +  y  =  4. 
(6)    x~  +  y2  =  25  and  3x-4t/=25. 
(C)    ar  +  y2  =  25  and  x  +  2  y  =  10. 

(d)  3.i-9  +  4?/2  =  24  and  a2-?/2  =  4. 

(e)  y2  =  4:X  and  #  —  ?/  -f- 1  =  0. 
(/)  a-9-f  4?/2  =  16  and  6y  =  x2. 

3.  If  the  equations  of  the  sides  of  a  triangle  are  x  +  7  y  + 11 
=  0,  3 x  +  ?/  —  7  =  0,  and  #  —  3y  +  l  =  0,  find  the  length  of 
each  of  the  medians. 


Ch.  Ill,  §  21]  LOCI  33 

4.  Which  of  the  points  (3,  -1),  (7,  2),  (0,  -2),  and  (8,  ,3) 
are  on  the  locus  of  the  equation  4  x  —  7  y  =  14. 

5.  Find  the  length  of  the  chord  of  intersection  of  the  loci 
of  7?  +  y-  =  13  and  y-  =  3  x  4-  3. 

6.  For  what  values  of  b  are  the  two  intersections  of  the 
loci  of  y  =  2  x  +  b  and  \f  =  4  a;  real  and  distinct  ?  imaginary  ? 
coincident  ? 

7.  Write  a  single  equation  which  will  represent  the  two 
bisectors  of  the  angles  between  the  axes. 

8.  Plot  the  two  lines  which  are  represented  by  each  of  the 
following  equations : 

(a)  x2  +  xy  =  0.  (c)    2x>  4-  5xy  -  3y2  =  0. 

(b)  x2-ox  =  -6.  (d)  2if-xy  +  ±x-9y  =  -4:. 


CHAPTER    IV 


THE    STRAIGHT   LINE 


22.  We  have  seen  that,  if  we  know  the  law  of  the  move- 
ment of  a  point,  we  can  often  determine  the  equation  of 
its  locus.  We  shall  now  proceed  to  the  systematic  study 
of  a  few  such  loci,  beginning  with  the  straight  line. 

The  two  most  common  ways  of  determining  the  position 
of  a  line  are  to  give  either  two  points  on  it,  or  a  single 
point  and  the  direction  of  the  line.  If  either  of  these  sets 
of  conditions  is  given,  the  line  is  fully  determined,  and  we 
should  be  able  to  find  the  algebraic  relation  which  must  be 
satisfied  by  every  point  on  it. 

23.  Line  through  two  points.  —  Let  the  line  pass  through 
the  two  points  Pv  (xv  y^),  and  P2>  (z2,  y2),  and  let  P, 
(#,  y),  be  any  point  on  the  line.  Draw  the  or  din  ate  s 
3I1PV  MP,  and  M2PV  and  the  line  PXK  parallel  to  OX. 
Then  from  the  similarity  of  the  two  triangles  PXLP  and 

Y 


P1KP2  we  have 
LP      KP 


2.. 


Fig.  27. 


PXL     PXK 

But  LP  =  y-  yx 
PXL  =  x  —  xv 
KP2  =  y2-  yv 
P1K=x2  -xv 


34 


Ch.  IV.  §23] 


THE   STRAIGHT    LINE 


35 


Substituting  these  values,  we  have 

y  -V\  =  Vi-y\t  1-5-1 

This  is  then  the  algebraic  relation  between  the  coordi- 
nates x  and  y  of  any  point  on  the  line  and  the  constants 
xv  yv  x2,  and  y2,  and  is  therefore  the  equation  of  the 
line.  It  is  called  the  two-point  form  of  the  equation  of 
the  straight  line. 

Let  the  student  show 
that  this  equation  cannot 
be  satisfied  by  the  coor- 
dinates of  any  point  not 
on  the  line. 

The  student  should 
here,  and  in  all  the  fol- 
lowing demonstrations, 
assure  himself  that  the 
proof  is  perfectly  general.  Place  the  lines  and  points  in 
different  positions,  being  careful  to  give  the  same  letter 
to  corresponding  points,  and  the  demonstrations  ought  to 
hold,  letter  for  letter.  For  example,  try  Fig.  28  with  the 
above  demonstration,  being  careful  to  note  that 

PXL  =  Mx0  +  OM    =  OM    -  0MV 
LP    =  LM  +  MP    =MP    -  ML, 
P1K=3I10-{-  0M2  =  0M2   -  0MV 


KP,  =  KM,  +  M9PK 


M2P2  -  M2K. 


Equation  [5]  may  be  written  in  the  determinate  form 
x%    y      1 


=  o. 


3G  ANALYTIC   GEOMETRY        [Ch.  IV,  §§  24,  25 

24.  Line  determined  by  its  intercepts.  —  If  the  two 
given  points  should  be,  in  particular,  the  points  where  the 
line  cuts  the  axes,  or  if,  in  other  words,  the  intercepts  a 
and  b  are  given,  the  equation  can  be  found  easily  by  sub- 
stituting («,  0)  for  (xv  y^)  and  (0,  5)  for  (#2,  y^)  in 
equation  [5].     It  becomes 

y  _  Q  =  b  -  0 
x  —  a      0  —  a 

or  reducing,  — +  ?  =  1.  [6] 

This  is  called  the  intercept  form  of  the  equation  of  the 
straight  line. 

.  Let  the  student  derive  equation  [6]  geometrically  with- 
out using  equation  [5]. 

25.  Oblique  coordinates. — In  obtaining  these  equations 
of  the  straight  line  we  have  made  no  use  of  the  fact  that 
the  axes  are  perpendicular.  The  011I3-  idea  used  was  the 
similarity  of  triangles,  which  will  be  true  in  oblique  as 
well  as  rectangular  coordinates.  The  results  will  hold 
therefore  for  both  systems  of  Cartesian  coordinates. 

PROBLEMS 

1.  Find  the  equation  of  the  straight  line  through  the  points 
(-  1,  5)  and  (6,  0). 

Solution.  —  In  applying  formula  [5]  either  point  may  be  chosen  as 
Pi  and  the  other  as  P2.  Here  let  (6,  0)  be  Pi  and  (-  1,  5)  be  P2.  Sub- 
stituting in  [5],  we  have  as  the  equation  of  the  line  5  x  +  7  y  =  30. 

2.  Find  the  equations  of  the  lines  through  the  following 
points  and  find  the  intercepts  of  these  lines  on  the  axes : 

(a)    (-  5,  4)  and  (S,  -  1).  (c)    (4,  2)  and  (4,  -  2). 

(6)    (0;  0)        and  (4,  3).  (d)  (3,  5)  and  (-  7,  5). 


Ch.  IV,  §  26] 


THE    STRAIGHT   LINE 


37 


3.  Find  the  equation  of  the  line  whose  intercepts  are  3 
and  —  1. 

4.  Does  the  line  joining  the  two  points  (6,  0)  and  (0,  4)  pass 
through  the  point  (3,  2)  ?  the  point  (—4,  5)  ? 

5.  What  condition  must  be  satisfied  if  the  point  (xlf  y{)  lies 
on  the  line  joining  the  points  (x2,  y.,)  and  (x3y  y3)  ? 

6.  The  line  joining  the  points  (6,  2)  and  (7,  —  3)  is  divided 
in  the  ratio  of  2  to  5.  Find  the  equation  of  the  line  joining 
the  point  (—  5,  —  5)  to  the  point  of  division. 

7.  The  coordinates  of  the  vertices  of  a  triangle  are  (2,  1), 
(3,  —2),  and  (—4,  —  1).  Find  the  equation  of  the  medians, 
and  show  that  the  coordinates  of  the  point  of  intersection  of 
any  two  medians  satisfy  the  equation  of  the  third,  and  that 
the  three  medians  therefore  meet  in  a  point. 

8.  What  are  the  equations  of  the  diagonals  of  the  rectangle 
whose  vertices  are  (0,  0),  (a,  0),  (0,  6),  and  (a,  b)  ?  Find  the 
point  of  intersection,  and  show  that  they  bisect  each  other. 

9.  What  system  of  lines  is  represented  by  the  equation 

y 


X 

-+. 

a     o 


1,  if  we  keep  a  constant  and  allow  b  to  vary  ?   if  we 

keep  b  constant  and  allow 
a  to  vary  ? 

26.  Line  determined  by 
a  point  and  its  direction. 
—  If  the  second  condi- 
tion mentioned  in  Art. 
22  be  given,  —  a  point  on 
the  line  and  the  direc- 
tion of  the  line,  —  we  can 
obtain  its  equation  as 
follows : 

Let  (xv  y{)  be  the  given  point,  and  let  the  direction  of 
the  line  be  determined  by  the  angle  <y  which  it  makes  with 


Fig.  29. 


38  ANALYTIC   GEOMETRY  [Ch.  IV,  §  27 

the  positive  direction  of  the  X-axis  measured  in  the  posi- 
tive direction  of  rotation.      In  the  triangle  KPXP, 

KP 


P1iT=UU'^ 

But  for  all  positions  of  P, 

KP  =  y-yv 

PXK=  x  —  xv 

and 

tan  KPXP  =  tan  7. 

Hence 

y  ~y^  =  tan  7, 
x  —  x1 

or  y  -  yl  =  t'&nv(x-xl). 

Tan  7  is  called  the  slope  of  the  line,  and  may  be  repre- 
sented by  I.     The  equation 

y-  2/1  =  i(oc-xi)  [7] 

is  called  the  slope-point  form  of  the  equation  of  a  line. 
By  comparing  equations  [5]  and  [7]  we  see  that 

xi    x\ 

27.    Line  determined  by  its  slope  and  its  intercept  on  the 

X-axis.  —  If    the    point    through    which    the    line    is    to 

pass  lies  on  the   Y-axis,  its  coordinates  being  (0,  5),  [7] 

reduces  to 

y  -b  =  Ix, 

or  y  =  lx  +  b.  [8] 

This  is  called  the  slope  form  of  the  equation  of  a  line. 
Let    the    student    derive    equation    [8]    geometrically 
without  using  equation   [7]. 


Ch.  IV,  §  28] 


T1IK    STRAIGHT    LINE 


39 


28.  Oblique  coordinates.  —  In  deriving  equations  [7] 
and  [8]  we  have  made  use  of  the  fact  that  the  axes  are 
rectangular.  A  separate  demonstration  is  therefore 
necessary  in  oblique  coordinates. 

Using  the  same  construction  and  notation  as  in  Art.  26, 
it  is  again  true  that  PXK  =  x  —  xx  and  KP  =  y  —  yv  But 
the  triangle  KPXP  is 
not  right-angled,  and  in 
order  to  find  the  ratio 
between  its  sides  we 
must  make  use  of  the 
law  of  the  sines.  Let 
the  positive  direction  of 
PXP  be  taken  along  the 
terminal  line  of  the  an- 
gle 7.  Then  the  angle 
formed  by  the  positive 

direction  of  PXP  with  the  positive  direction  of  PXK  is 
always  7,  and  the  angle  formed  by  the  positive  direction 
of  ifi^  with   the  positive  direction  of  PXP  is   (60—7). 


Fig.  30. 


Hence, 


K?-  =  y=Jh=   .    si°7  (See  Art.  7) 

Px K       x  —  xx      sin  (o>  —  7) 


or 


sni  («  —  7) 


[9] 


If  the  coordinates  of  the   given  point  are  (0,  b),  [9] 
reduces  to 

y=  .  ^   ^+&-  [10] 

Sill  (ft)  -  y)  L       J 

When  ay  =  90°,  these  two  forms  will  be  seen  to  reduce 
to  the  equations  [7]  and  [8]. 


40  ANALYTIC   GEOMETRY  [Ch.  IV,  §  29 

PROBLEMS 

1.  What  is  the  equation  of  the  line  which  passes  through 
the  point  (—6,  6)  and  makes  an  angle  of  60°  with  the  X-axis  ? 

2.  Find  the  equation  of  a  straight  line  if 

(a)  b  =  6  and  y  =  30°, 

(b)  b  =  -5  and  y  =  tan"1  f , 

(c)  b  =  S,   y  =  30°,    and  <o  =  00°. 

3.  Find  the  equation  of  the  straight  line  through  the  inter- 
section of  the  lines  2  x  —  3  y  =  4  and  3  x  —  y  =  o,  and  making 
an  angle  of  120°  with  the  X-axis. 

4.  What  is  the  slope  of  the  line  whose  intercept,  on  the 
Y"-axis  is  5  and  which  passes  through  the  point  (3,  —  1)  ? 

5.  What  system  of  lines  is  represented  by  the  equation 
y  =  lx  +  b  if  we  keep  I  constant  and  allow  b  to  vary  ?  if  we 
keep  b  constant  and  allow  I  to  vary  ? 

29.  General  equation  of  the  first  degree.  —  We  have 
found  that  the  equation  of  every  straight  line  given  by 
any  of  the  preceding  conditions  is  of.  the  first  degree 
in  both  rectangular  and  oblique  coordinates.  It  now 
remains  to  consider  whether  an  equation  of  the^iirst 
degree  can  represent  any  other  locus. 

Every  such  equation  is  included  in  the  general  form 

where  A,  B,  and  C  can  have  any  values,  positive,  negative, 
or  zero. 

If  B^O,  we  can  divide  the  equation  by  it,  and  trans- 
posing, we  have 

A         C 
u  B         B 


THE   STRAIGHT    LINK 


n 


Ch.  IV,  §  20] 

A  C 

where   — —  and  — -  can  liave  any  value.      But  the  slope 

form   of  the   equation   of  a  line   has  been   shown   to    be 

y  =  lx  +  b.  (Arts.   27,  28) 

Since  1=  tan  7     or  in  oblique  coordinates  sm  ^ — 

L  sin(ft>-7)J 

and  b  is  the  intercept  on  the  !F-axis,  they  can  have  any 
real  value  whatever. 

AVe  have  then  reduced  the  general  equation 
Ax  +  By  +  O  =  0 
to  the  slope  form  of  the  equation  of  a  line,  and  it  must 
represent  that  line  for  which 


1  = 


and   b =  — 


If  B 


B  B 

0,  the  general  equation  reduces  at  once  to 

a 


X  =  — 


which  we  know  to  be  the  equation  of  a  line  parallel  to  the 
y-axis. 

We  have  then  shown 
that  the  general  equation 
of  the  first  degree  always 
represents  a  straight  line. 

Another     method     of 
showing    that    the   locus   ■x_ 
of   any  equation  of   the 
first  degree  is  a  straight 
line  is  as  follows  :  ■FlG*     • 

Let  (xv  ?/1),  (xT  y2),  and  (#3,  y%)  be  the  coordinates  of 
any  three  points  on  the  locus  of  the  equation 
Az  +  By  +  C=Q, 


42  ANALYTIC    GEOMETRY  [Ch.  IV,  §  30 

These  coordinates  must  satisfy  the  equation.     Substi 
tuting,  we  have 

(1)  Ax1  +  By1  +  C=Q, 

(2)  Ax2  +  By2  +  C=0, 

(3)  ^%+%8  +  (7=0. 
Subtracting  (2)  from  (1),  we  have 

-A(x2-  xx)  =  B(y2-  yi), 
xn  —  x,      B 


or 


i  - 


y\-y<i    A 

Subtracting  (3)  from  (2),  we  have 

-  A  (xz  -  z2)  =  B  (j/3  -  y2), 


or 


•T3  -  X2  _  B. 

Vi-y*    A 

x2  - 

1  =  -2 2,   or  2  : 

- 1/2    y%- y*      KI\ 

LP^ 
LP2 

Hence, 


The  triangles  KPXP2  and  LP2PZ  are  therefore  similar, 
and  P1P2P3  is  a  straight  line. 

30.  Two  equations  representing  the  same  line  cannot 
differ  except  by  a  constant  factor.  — Let  A1x  +  B1y  +  Cx  =  0 
and  A2x  +  B2y  +  02  =  0  represent  the  same  line.  Their 
intercepts  on  the  axes  must  be  the  same.     Hence, 


and 


4= 

2x  = 

Bx 

_4 

"3" 

Ch.  IV,  §  30]  THE    STRAIGHT   LINE  43 

ABC 

Hence,  — 1  =  — 1  =  -J,  and  the  equations  differ  only  by  a 

A2      -E>2      *  2 

constant  factor. 

The  converse  is  easily  seen  to  be  true  :  if  tiro  equation* 
of  the  first  degree  differ  only  by  a  constant  factor^  they  rep- 
resent the  same  straight  line, 

PROBLEMS 

1.  Find  the  values  of  a,  b,  and  I  for  the  line  whose  equation 
is  2<c  +  3y-12  =  0. 

Solution.  —  The  intercepts  are  found,  as  explained  in  Art.  19,  to  be 
a  —  6  and  b  =  4.  The  slope  I  may  be  found  by  changing  the  equation 
into  the  slope  form,  as  explained  in  Art.  29.  Transposing  and  dividing 
by  3,  we  have  y  =  —  §  x  +  4.     Hence  I  =  —  §. 

2.  Find  the  values  of  a,  b,  and  I  for  the  lines  represented 
by  the  following  equations,  and  construct  the  lines  first  by  the 
aid  of  the  intercepts  a  and  b,  then  by  the  aid  of  the  slope  I, 
and  the  intercept  b  : 

(a)      x  —  ±y  — 10  =  0,  (c)   4#  +  #  =  0, 

(5)  3a_5<?/+    7  =  o,  (d)  2^  +  8  =  0. 

3.  Determine  the  values  of  A,  B,  and  C,  if  the  line 
Ax  +  By  -f-  C  =  0  passes  through  the  points  (3,  0)  and  (2,  —  1). 

Solution'.  —  Since  the  line  is  to  pass  through  these  points,  their  coordi- 
nates must  satisfy  its  equation.     By  substitution,  we  obtain 

3.4+  (7  =  0, 

and  2A-B+C=Q, 

two  equations  in  A,  B,  and  C,  from  which  the  values  of  two  of  them  may 
be  obtained  in  terms  of  the  third.  Solving,  we  have  C  =  —  3  A  and 
B  =  —  A.     The  equation  of  the  line  is  therefore 

Ax  -  Ay  -  3  A  =  0, 

cr  x  -  y  -  3  =  0. 


44 


ANALYTIC    GEOMETRY 


[Ch.  IV,  §  31 


4.  Find  by  the  same  method  the  equations  of  the  lines 
through  the  points 

(a)  (3,  1)  and  (-5,0), 

(b)  (0,-2)  and  (3,4), 

(c)  (0,  0)  and  (5,  -  3). 

5.  Show  that  if  two  lines  are  parallel,  their  slopes  must  be 
equal ;  if  perpendicular,  the  slope  of  one  must  be  the  negative 
reciprocal  of  the  slope  of  the  other,      [tan  y  =  —  cot(y  +  90°).] 

6.  Select  pairs  of  the  following  equations  which  represent 
(a)  parallel  lines,  (6)  perpendicular  lines : 

2x-3y  =  6,  x  =  -$y  +  6, 


4  x  —  6  y  =  7, 
12aj  +  8y  =  ll. 


31.  The  angle  which  one  line  makes  with  another.  — We 
have  defined  the  angle  between  two  directed  lines  as  the 
angle  between  their  positive  directions.  But  when  the 
lines  are  given  by  their  equations  and  no  convention  is 

used  to  fix  their  posi- 
tive directions,  it  is  con- 
venient to  define  the  an- 
gle which  one  line  makes 
with  another  as  the  angle 
formed  by  going  from  the 
second  line  to  the  first  in 
the  positive  direction  of 
rotation.  This  definition 
always  gives  a  definite 
angle. 

In  Fig.  32,  the  angle  which  AB  makes  with  MJV  is 
Z  NOB,  or  its  equal  Z  MO  A  ;  the  angle  which  MN  makes 
with  AB  is  Z  BOM  or  Z.  AON.     Let  it  be  required  to 


Ch.  IV,  §31]  THE   STRAIGHT  LINK  45 

find  the  angle  which  AB  makes  with  MN.     Let  the  equa- 
tions of  AB  and  MN  be  given  in  the  form 

y  =  hx  +  hv 

and  y  =  l2x  +  b2. 

From  the  figure  #  =  y{  —  y2, 

and  tan  g  =  tan  (7. -7.)  =    tan  7,  -  tan  7,  , 

1  +  tan  yx  tan  y2 

Hence  tan0  =  ^f-^-.  [11,  a] 

If  the  equations  are  given  in  the  form 
A1x  +  B1y-{-C1  =  0, 
and  A2x  +  i?2?/  -f  C2  =  0, 

it  was  shown  in  Art.  29  that 

*i  =  -  -g1  and  h=~jf' 

PROBLEMS 

1.  Find  the  angle  which  the  line  5x  —  3y  =  10  makes  with 
the  line  x  -f-  2  ?/  =  7. 

Solution. — The  angle  is  to  be  measured  from  the  last  line,  and  that 
line  therefore  takes  the  place  of  MN  in  Fig.  32.  Hence  h  =  f,  and 
J. 2  —  —  £.     Substituting  these  values  in  [11,  a], 

tan  0  =  i-±i  =  13, 

1—5  ' 

1  6 

or  6  =  tan"1 13. 

If  the  question  is  reversed,  and  we  wish  to  find  the  angle  which  the 
line  x  +  2  y  =  7  makes  with  the  line  5  x  —  3  ?/  =  10,  we  must  take  h  =  —  \ 
and  h  =  f .     Then 

tan  6  =  ~  ^  ~  I  =  _  13 
1-1 
or  0  =  tan-1  (-  13). 


40  ANALYTIC   GEOMETRY  [Ch.  IV,  §  32 

2.  Find  the  angle  which  the  line  3x-\-oy  —  l  =  0  makes 
with  the  line  11  a?  —  2y  +  3  =  0. 

3.  Find  the  interior  angles  of  the  quadrilateral  whose  ver- 
tices are  (3,  3),  (5,  -  3),  (4,  -  5),  and  (-  3,  0). 

4.  The  equations  of  the  sides  of  a  triangle  are  #+8  y  +  11  =  0, 
2  x  —  3  y  +  1  =  0,  and  4  x  +  5  ?/  +  6  =  0.  Find  one  exterior 
angle  of  the  triangle  and  the  two  opposite  interior  angles. 

32.  Perpendicular  and  parallel  lines.  —  If  two  lines  are 
parallel,  tan  6  =  0,  and  therefore  lx  —  l2  =  0,  or 

This  appears  also  from  the  figure,  since  if  two  lines  are 
parallel,  they  must  make  the  same  angle  with  the  X-axis. 

If  two  lines  are  perpendicular,  tan  6  =  oo,  and  therefore 
1  +  lj2  =  0,  or 

This  appears  also  from  the  figure,  since  if  the  lines  are 
perpendicular, 

7i  =  72  -  |-i  and    tan  7i  =  -  cot  72<  or  ?i  =  ~  y' 

If  now  we  wish  to  obtain  the  equation  of  a  line  parallel 
to  a  given  line  Ax  +  By  +  C  =  0,  the  only  condition  which 
must  be  satisfied  is  that  it  shall  have  the  same  slope. 
This  can  be  accomplished  by  writing  the  equation 
Ax  +  By  =  &,  where  h  is  arbitrary.  This  will  include  all 
lines  parallel  to  the  given  line,  for  by  varying  k  it  can 
be  made  to  represent  any  one  of  the  indefinite  number 
of  such  lines.  The  value  of  k  in  any  particular  problem 
must  be  determined  by  some  other  condition.  For 
example,  if  it  is  to  pass  through  a  given  point,  k  can  be 


Ch.  IV,  §  32]  THE   STRAIGHT   LINE  47 

determined  from  the  fact  that  the  coordinates  of  this 
point  must  satisfy  the  equation. 

Again,  if  we  wish  to  obtain  the  equation  of  a  line  per- 
pendicular to  the  line  Ax  +  By  +  C  =  0,  we  must  write 

A  B 

an  equation  such  that   — l  = -.     Such    an  equation   is 

JDj  A 

Bx  —  Ay  =  k.  This  again  contains  all  the  lines  perpen- 
dicular to  the  given  line.  If  the  line  is  to  pass  through 
a  given  point,  k  can  be  determined  as  before  by  the  fact 
that  the  coordinates  of  this  point  must  satisfy  the 
equation. 

PROBLEMS 

1.  Write  the  equations  of  the  lines  through  (3,  4)  which 
are  respectively  parallel  and  perpendicular  to  the  line 
3  x  —  5  y  =  10. 

Solution.  — The  equation  of  the  line  which  is  parallel  will  be  of  the 
form  3  x  —  5  y  =  k.  Substituting  (3,  4),  we  have  k  =  — 11,  and  the  equa- 
tion of  the  parallel  line  is  3  x  —  by  =  —  11. 

The  equation  of  the  perpendicular  line  will  be  of  the  form  bx  +  Sy  =  k. 
Substituting  (3,  4),  k  =  27,  and  the  equation  of  the  perpendicular  line  is 
bx-\-  Sy  =  21. 

2.  Find  the  equation  of  the  line  through  (5,  8)  perpendicular 
to  3  x  +  7  y  =  21. 

3.  In  the  triangle  whose  vertices  are  (0,  0),  (6,  0),  and  (4,  8), 
find  (a)  the  equations  of  its  sides ;  (b)  the  equations  of  perpen- 
diculars from  the  vertices  upon  the  opposite  sides ;  (c)  the 
equations  of  the  perpendicular  bisectors  of  the  sides ;  (d)  the 
equations  of  the  medians. 

4.  Show  that  in  the  above  problem  the  perpendiculars  from 
the  vertices,  the  perpendicular  bisectors,  and  the  medians  each 
meet  in  a  point. 

5.  Show  that  the  points  obtained  in  problem  4  lie  on  a  line, 
and  obtain  the  ratio  of  the  distances  between  them. 

6.  Show  that  in  any  triangle  the  medians  meet  in  a  point. 


48 


ANALYTIC    GEOMETRY 


[Ch.  IV,  §  33 


Note. — Choose  the  axes  of  coordinates  so  that  the  origin  is  at  one 
vertex  and  the  X-axis  is  coincident  with  one  side  of  the  triangle.  The 
coordinates  of  the  vertices  of  the  triangle  may  then  be  taken  as  (0,  0), 
(a,  0),  and  (&,  c). 

7.  Show  that  in  any  triangle  the  perpendiculars  from  the 
vertices  on  the  opposite  sides  meet  in  a  point. 

8.  Show  that  in  any  triangle  the  perpendicular  bisectors  of 
the  sides  meet  in  a  point. 

9.  Show  that  the  three  points  obtained  in  problems  6,  7, 
and  8  lie  on  a  line,  and  find  the  ratio  of  their  distances  from 
each  other. 

10.  Show  that  the  line  joining  the  middle  points  of  two 
sides  of  a  triangle  is  parallel  to  the  third  side  and  equal  to  one 
half  of  it. 

11.  Show  that  the  diagonals  of  a  square  or  rhombus  are 
perpendicular  to  each  other. 


33.    Line  making  a  given  angle  with  a  given  line.  — In 

plane  geometry  it  is  usual  to  speak  of  two  lines  through 

any  given  point  and  making  a  given  angle  with  a  given 

line.  But  if  we  con- 
sider the  direction  of 
the  angle,  there  can  be 
only  one  such  line.  For 
if  MN  is  the  given  line, 
Pj  the  given  point,  and 
(f>  the  given  angle,  there 
can  be  only  a  single  line 
which  passes  through  Px 
and  makes  the  angle  <j> 

with  MN,  where  </>  is  measured  in  the  positive  direction  of 

rotation. 


Ch.  IV,  §  34]  THE   STRAIGHT  LINE  49 

Let  ES  be  this  line.  Let  the  inclination  of  MN  be  yv 
and  of  BS  be  7.     Then  from  [11,  a], 

tan  d>  = —L. 

*      1  +  «, 

Solving  for  7,  we  have 

,  _   /1  4-  tan  (f) 
1  —  Zj  tan  (/> 

The  equation  of  BS  will  therefore  be 

If  MS  is  parallel  to  J£ZV",  tan  <f>  =  0,  and  the  equation 
becomes 

If  BS  is  perpendicular  to  ilifiV,  tan  <f>  =  go,  and  the 
equation  becomes 

l\ 
These  formulas  might  be  used  to  write  the  equations  of 
parallels  and  perpendiculars  in  place  of  the  methods  given 
in  the  previous  section. 

PROBLEMS 

1.  Find  the  equation  of  the  line  through  the  origin  which 
makes  an  angle  of  60°  with  the  line  x  —  3  y  =  10. 

2.  Find  the  equation  of  the  line  through  (1,  4)  which  makes 
an  angle  of  135°  with  the  line  joining  (1,  4)  with  the  intersec- 
tion of  5  x  —  2  y  =  17  and  3  x  +  4  y  =  5. 

34.  Normal  form  of  the  equation  of  a  straight  line.  — 
If  we  have  given  the  length  of  the  perpendicular  or  nor- 
mal from  the  origin  on  a  line,  together  with   the  angle 


50 


ANALYTIC    GEOMETRY 


[Ch.  IV,  §  34 


which  this  normal  makes  with  the  positive  direction  of  the 
.X-axis,  the  line  is  completely  determined.  The  perpen- 
dicular distance  is  represented  by  jt?,  and  the  angle  by  a. 

Through  0  draw  a  line 
making  an  angle  a  with 
OX.  If  any  distance  Off 
is  laid  off  on  this  line 
either  in  the  positive  di- 
rection (along  the  termi- 
nal line  of  the  angle),  or 
in  the  negative  direction, 
and  through  H  a  line 
AB,  perpendicular  to  Off, 
is  drawn,  that  line  is  com- 
pletely determined.  It 
is  convenient  to  restrict  a  to  positive  values  from  0°  to 
360°.  In  case  we  wish  to  speak  of  a  complete  set  of  parallel 
lines  without  changing  «,  it  will  be  necessary  to  allow  p  to 
be  either  positive  or  negative,  but  every  line  in  the  plane 
can  be  determined  by  positive  values  of  both  a  and  p,  and 
this  will  always  be  understood  unless  otherwise  stated. 
We  have  seen  that  the  equation  of  the  line  AB  in  terms 

of  its  intercepts  is  -  +  %-  =  1. 

a     o 
line 


Fig.  34. 


But  for  all  positions  of  the 


V 


P 


and 


-=cos«, 

or 

a  = 

cos  a 

p 

f-  =  Sill  «, 
0 

or 

sin  a 

Substituting  these  values  of  a  and  5,  the  equation  of  AB 

becomes 

x  cos  a  +  y  sin  a  =  p.  [15] 


Ch.  IV,  §  35]  THE   STRAIGHT  LINE  51 

This  is  called  the  normal  form  of  the  equation  of  a 
straight  line. 

Let  the  student  show  that  the  equation  of  a  straight 
line  in  oblique  coordinates  in  terms  of  a  and  p  is 

x  cos  a  -f-  y  cos  (o>  —  a)  =  p. 

Note.  —  The  equations  -  =  cos  a  and    -  =  sin  a  are  true  for  all  cases, 
a  b  ' 

since  if  p  is  positive,  a  and  cos  a  have  the  same  sign,  and  also  b  and  sin  a. 
While  if  p  is  negative,  they  have  the  opposite  signs. 


PROBLEMS 
1.    What  is  the  equation  of  the  straight  line  in  which 
(a)  a  =    60°,  and  p  =  5  ?       (d)  a  =  225°,  and  p  =  0  ? 
(6)  a  =  120°,  and  j>  =  5  ?       (e)  a  =  45°,  u>  =  60°,  and  p  =  1  ? 
(c)   « = 330°,  and  p=  - 5  ?      (/)  a=  - 60°,  <o = 135°,  and  p  =  6  ? 

35.  Reduction  of  the  general  equation  to  the  normal 
form.  —  Since  the  general  equation  of  the  first  degree 
Ax  H-  By  -f  O  =  0  always  represents  a  straight  line,  it 
ought  to  be  possible  to  reduce  it  to  any  one  of  the  stand- 
ard forms.  We  have  already  shown  how  to  reduce  it  to 
the  slope  form,  and  that 

I  =  —  — ,  and  b  =  —  —  • 
B  B 

The  following  method  enables  us  to  reduce  it  to  the 
normal  form.  If  the  two  equations  Ax  +  By  +  C  =  0 
and  x  cos  a  +  y  sin  a  —  p  =  0  are  to  represent  the  same 
line,  it  was  shown  in  Art.  30  that  they  can  differ  only  by 
a  constant  factor.     Let  k  be  the  quantity  by  which  it  is 


52  ANALYTIC   GEOMETRY  [Ch.  IV,  §  35 

necessary  to  multiply  Ax  +  By  -f-  C=  0  to  make  it  iden- 
tical with  x  cos  a  -f-  y  sin  a  —  p  =  0. 

Then     Ar^L  =  cos  «,  &Z?  =  sin  «,  and  kO=  —  p. 

Squaring  the  first  two  and  adding,  we  have 

k2A2  +  k2B*  =  cos2  a  +  sin2  a  =  1. 

Hence      &  =  ±  —  .   and  the  equation 

VA2  +  B2 

A         v  + B         y  + C         =0      [16] 

±VA*  +  B*  ±y/A*  +  B*  ±y/A*  +  Bf 

is  then  identical  with  x  cos  a  +  y  sin  a  —  p  =  O! 

If  a  and  jt?  are  so  chosen  that  ^>  shall  always  be  positive, 
then  in  any  numerical  case  that  sign  must  be  given  to  the 

C 
radical  which  will  make .     a  negative  number 

±  VA2  +  B2 
to  correspond  to  —  p.     Hence  the  sign  of  the  radical  mast 
be  chosen  opposite  to  the  sign  of  O.     This  will  always  be 
understood  unless  the  contrary  is  stated. 

PROBLEMS 

1.  Reduce  the  equation  3  x  -f-  4  y  =  10  to  the  normal  form. 

Solution.  —  Here  ±  VA2  +  B2  =  ±  5,  and  since  C  is  negative,  we  must 
divide  by  +  5,  and  the  equation  becomes  f  x  +  $  y  =  2. 

Hence  cos  a  =  f,  sin  a  =  f,  and  p  =  2.  The  line  can  be  easily  plotted. 
What  would  have  been  the  values  of  a  and  p,  if  —  5  had  been  chosen  ? 

2.  Reduce  the  following  equations  to  the  normal  form  and 
plot  the  lines  which  they  represent : 

(a)  lx-3y  =  25,  (d)  aj  +  4  =  0, 

(6)  ».  +  2y  =  -8,  (e)   5y-3  =  0, 

(c)  2x-2/  =  0,  (/)  aj-3y  +  4  =  0. 


Ch.  IV,  §  36] 


THE    STRAIGHT   LINE 


53 


3.  What  system  of  lines  is  represented  by  the  equation 
x  cos  a  -f  y  sin  a  —  p  =  0,  if  we  keep  a  constant  and  allow  p  to 
vary  ?     If  we  keep  j>  constant  and  allow  a  to  vary  ? 

36.  Distance  of  a  point  from  a  line.  —  Let  it  be  required 
to  find  the  distance  of  the  point  P1  from  the  line  AB  when 
the  equation  of  AB  is  given  in  the  form 

x  cos  a  +  y  sin  a  —  p  =  0. 


Draw  MJST  through  Pl  parallel  to  AB  and  continue  the 
perpendicular  OH  to  meet  it  at  K.  The  equation  of  MN 
will  be 

x  cos  a  +  y  sin  a  —  p1  =  0, 

where  px  may  be  either  positive  or  negative.  For,  as  the 
value  of  a  is  fixed  and  as  MN  can  be  any  line  parallel  to 
AB,  it  may  be  on  the  opposite  side  of  the  origin  from  AB, 
and  in  this  case  px  will  be  negative.     (See  Art.  34.) 

Since  Px  lies  on  MN,  its  coordinates  (xv  y-^)  must  satisfy 
the  equation  of  MN, 


54  ANALYTIC   GEOMETRY  [Ch.  IV,  §  36 

Hence  xx  cos  a  +  yx  sin  a  =  pv 

Now  wherever  Px  may  lie,  RPX  =  HK=  OK-  011=  px-  p. 

Hence  BPi  =  a?icosa  +  yisma-p,  [17,  a] 

If  the  equation  is  given  in  the  form  Ax  +  By  +(7=0, 
it  is  necessary  first  to  reduce  it  to  the  normal  form  and 
then  substitute  xx  for  x  and  yx  for  y. 

Hence  MP,  =    ±VJFTW  •  [IT,  8] 

TAe  radical  must  be  given  the  sign  opposite  to  that  of  C. 

It  appears  from  the  way  RPX  has  been  chosen  that  the 
result  will  be  positive  when  the  point  and  the  origin  are  on 
opposite  sides  of  the  line  ;  negative,  when  they  are  on  the 
same  side  of  the  line. 

PROBLEMS 

1.  Find   the    distance   of   the    point   (3,  5)   from    the   line 

2X-3y  +  6  =  Q. 

2.  Find  the  distance  of  the  origin  from  the  line 

3aj  +  4y-5  =  0. 

3.  Find  the  area  of  the  triangle  whose  vertices  are  (0,  3), 
(4,  0),  and  (5,  5)  by  calculating  the  length  of  one  side  and  the 
distance  of  the  opposite  vertex  from  that  side. 

4.  Given  the  line  Sx  —  4y  =  10  and  the  point  (—3,5). 
Find  the  equation  of  the  line  through  the  point  perpendicular 
to  the  given  line ;  find  the  point  of  intersection  of  this  perpen- 
dicular with  the  given  line ;  find  the  distance  of  the  given  point 
from  this  point  of  intersection. 

5.  Use  the  method  indicated  above  to  find  the  distance  from 
the  point  (a^,  yx)  to  the  line  Ax  +  By  +  C  =  0, 


Ch.  IV,  §  :J7J  THE    STRAIGHT   LINE  55 

6.  Find  the  distance  bet  ween  the  two  parallel  lines 

7  x  —  8  y  =  15,  and  7  x  —  8  y  -—  40. 
Which  line  is  nearer  the  origin? 

7.  Show  that  the  point  (3,  1)  is  on  the  same  side  of  the  line 
x  -+-  4  ?/  =  8  as  the  origin. 

37.  Oblique  coordinates.  —  It  will  be  noticed  that  sec- 
tions 31-36  have  reference  to  rectangular  coordinates 
only.  The  corresponding  formulas  in  oblique  coordinates 
are  rather  complicated  and  seldom  used.  We  shall  simply 
state  what  they  are  without  obtaining  them. 

To  reduce  Ax  -f  By  +  0 '=  0  to  the  normal  form, 
%  cos  a  -+-  y  cos  (to  —  a)  =  p,  multiply  the  equation  by 

sin  (o 


^A2  +  B*-2ABcosc0 


The  angle  between  two  lines  whose  equations  in  oblique 
coordinates  are  Alx  +  Bly  +  C1=0  and  A2x  -f  B2y  +  C2  =  0 
is 

tan  6  = (AXB2- A2B,~)  sin  <* 

AXA2  H-  BXB2  —  {A1B2  -f-  A^B^)  cos  <o 

The  condition  for  parallelism  is  the  same  as  in  rectangu- 

A       A 

lar  coordinates,  — 1  =  -— 2.     But  the  condition  for  perpen- 

v       i       ■         ■  l  2 

dicularity  is 

AXA2  +  BXB2  -  (AXB2  +  A2B{)  cos  co  =  0. 

If,  then,  only  parallel  lines  enter  into  a  problem,  oblique 
coordinates  may  be  used  with  advantage ;  but  if  it  is 
necessary  to  use  perpendicular  lines,  oblique  coordinates 
should  be  avoided. 


56 


ANALYTIC   GEOMETRY 


[Ch.  IV,  §  38 


The  equation  of  a  line  perpendicular  to  Ax  +  By  +  C=0 

(B  —  A  cos  a))  x  —  (A  —  B  cos  co)y  =  k. 
The  distance  of  a  point  from  a  line  is 

xx  cos  a  +  y1  cos  (ft)  —  «)  —  ^?, 
Q4.3?!  +  ByA  4-  (?)  sin  ft) 


or 


VA2  + .B2-  2ABcosa> 


A' 


/0 


38.    Bisector  of  the  angle  between  two  lines.  —  Let  the 
equations  of  the  two  lines  AB  and  MN  be 

(1)  Axx  +  ^  +  Cj  =  0, 

and 

(2)  ^  +  ##  +  C2  =  0, 

and  let  (V,  ?/)  be  any 
point  on  the  bisector  of 
the  angle  between  them. 
Since  every  point  in  the 
bisector  of  an  angle  is 
equally  distant  from  the 
sides,  HP'  and  KP1  are  numerically  equal.     But 

KPi  =  a  j  +  By  +  c1  and  Hpt  =  A0y  +  B0y  +  ^ 

±Vyl12+^12    '  ±Vvl22  +  ^22 

Hence  the  relation  which  must  exist  between  a;'  and  y 
in  order  that  P'  may  be  a  point  on  the  bisector  is 


Y' 
Fig.  37. 


Aix'  +  Bip'  +Ci_     A2x'  +  B2y'  +  C2( 
±  \/^ia  +  JSi2  ±  V.422T#? 


[18] 


If  the  signs  of  the  denominators  have  been  chosen  in 
accordance  with  the  rule  given  in  Art.  36,  the  positive 
siern  in  the   second   member  indicates    that   P'  and  the 


Ch.  IV,  §39]  THE   STRAIGHT   LINE  57 

origin  are  either  on  the  same  side  or  on  opposite  sides 
of  each  of  the  lines,  and  that  therefore  the  equation  repre- 
sents the  bisector  of  the  angle  in  which  the  origin  lies; 
while  if  the  minus  sign  is  chosen,  it  represents  the  bisector 
of  the  angle  in  which  the  origin  does  not  lie. 

If  either  Cx  or  C2  is  zero,  one  or  both  of  the  lines  pass 
through  the  origin,  and  this  test  cannot  be  used. 

PROBLEMS 

1.  Find  the  equations  of  the  bisectors  of  the  angles  between 
the  two  lines  3  x  —  4  y  =  10  and  4  x  -f  3  y  —  7.  Show  that  the 
two  bisectors  are  perpendicular. 

2.  Show  that  the  bisectors  of  any  pair  of  supplementary 
adjacent  angles  are  perpendicular  to  each  other,  using  the  two 
lines  in  Art.  38. 

3.  The  equations  of  the  sides  of  a  triangle  are  3a?  =  4y, 
4  x  =  —  3  y,  and  y  =  6.  Show  that  the  bisectors  of  the  interior 
angles  meet  in  a  point.  Show  also  that  the  bisector  of  the 
interior  angle  at  one  vertex  and  the  two  bisectors  of  the 
exterior  angles  at  the  other  vertices  meet  in  a  point. 

39.  Lines  through  the  intersection  of  two  given  lines.  — 
If  the  equations  of  two  given  lines  are 

(1)  Ap  +  Btf+0^% 

and  (2)  A2x  +  B$  -+-  02  =  0, 

and  we  form  the  equation 

(3)  Axx  +  Bxy  +CX  +  k(A2x^  +  B2y  +  C2)  =  0, 

where  k  can  have  any  value,  it  will  represent  for  every 
value  of  k  some  line  through  the  intersection  of  .the  first 
two.  For  the  coordinates  of  the  point  of  intersection 
of   the  loci  of  (1)  and  (2),  which  must  satisfy  both  of 


58  ANALYTIC   GEOMETRY  [Ch.  IV,  §  39 

these  equations,  must  satisfy  (3)  also.  Moreover,  it 
represents  any  line  through  their  intersection,  for  k 
can  always  be  chosen  so  as  to  make  the  locus  of  (3) 
pass  through  any  given  point.  It  is  only  necessary  to 
substitute  the  coordinates  of  the  point  in  the  equation 
and  determine  k  so  that  the  equation  is  satisfied.  In 
this  wa}r  the  equation  of  the  line  through  the  intersection 
of  two  lines  and  any  other  point  may  be  obtained  without 
actually  finding  the  coordinates  of  the  point  of  intersection. 
If  any  other  condition  sufficient  to  determine  the  line 
is  given  (for  example,  its  slope),  k  can  always  be  deter- 
mined so  that  the  line  will  satisfy  the  condition. 

PROBLEMS 

1 .  What  is  the  equation  of  the  line  through  the  intersection 
of  2  x  +  3  y  -  4  =  0  and  x  +  2  y  -  5  =  0,  and  the  point  (2,  3)  ? 

Solution.  —  The  equation  of  any  line  through  the  intersection  of  the 
given  lines  is 

2x  +  3y-4  +  k(x  +  2y-  5)=  0. 

Since  the  line  is  to  pass  through  the  point  (2,  3),  these  coordinates 
must  satisfy  the  equation. 

Hence  k  =  —  3, 

Substituting  this  value,  we  have 

x  +  3y-ll=0, 
as  the  equation  desired. 

2.  What  is  the  equation  of   the  line  passing  through  the 
origin  and  the  intersection  of  the  lines  x  +  oy  —  8  =  0  and 

3.  In  the  triangle  whose  sides  are 

5x-6y  =  16,  4z  +  5?/=20,  and  x  +  2y  =  0, 

find  the  lines  through  the  vertices  and  parallel  to  the  opposite 
sides  without  finding  the  coordinates  of  the  vertices. 


Ch.  IV,  §  40] 


THE   STRAIGHT   LINE 


59 


4.  Find  the  equation  of  a  line  through  the  intersection  of 
the  lines  2x  —  3y  +  l  =  0  and  x  +  5  y  -f  6  =  0,  which  is  per- 
pendicular to  the  first  of  these  lines. 

5.  Find  the  equation  of  the  line  through  the  intersection 
of  the  lines  y  =  7  x  —  4  and  y  =  —  2  x  +  5,  which  makes  an 
angle  of  60°  with  the  X-axis. 

6.  Find  the  equation  of  the  line  through  the  intersection 
of  the  lines  5y  —  2x  — 10  =  0  and  y-f4a?  —  3  =  0,  and  also 
through  the  intersection  of  the  lines  10  y  -f-  x  -f-  21  =  0  and 
3y-5x  -fl  =  0. 

40.  Area  of  a  triangle.  —  If  the  coordinates  of  the 
vertices  of  a  triangle  are  given,  the  area  of  the  triangle 
may  be  found  in  the  fol- 
lowing manner : 

The   area   is  equal   to 
the  numerical  value  of 

\HP^PXPV 


Fig.  38. 


0. 


^2)2  +  (^l-y2)2- 
JIPS  is  the  distance  of    — 
P3  from  the  line  PXPT 
The  equation  of  PXP2  is 

02  -  Vi)x  ~  02  -  x\)y  -  *iy*  +  xtfx 

Hence  HP  =  ^2 ~ yi)x* ~ ^2 ~ X^y* ~ X^2  +  X^K 

and   the    area  =  \ [(y2-y{)xz ~(x%-x{)yz- xxy2 +a^yj, 

=  K  («i  -  ^2)2/3  +  (^2-^3)2/1  +  (a?3-a?i)2/2].      [19] 

The  form  of  the  result  is  easily  remembered  since  the 
subscripts  follow  the  cyclic  order.     The  sign  of  the  result 


60  ANALYTIC   GEOMETRY  [Cii.  IV,  §  40 

may  be  disregarded,  since  it  is  only  the  numerical  value 
of  the  area  we  wish. 

Formula  [19]  may  be  written  in  the  determinate  form 


A  =  i 


\ 


PROBLEMS 

1.  Find  the  area  of  the  triangle  whose  vertices  are  (1,  —  3), 
(-4,  3),  and  (5,5). 

2.  Show  that  the  area  of  any  quadrilateral  is 


[Oi?/2  -  a^/O  +  (x2y3  -  x3y2)  +  (x&4  -  x^)  +  (xAyx  -  xly,)\ 


3.  What  is  the  area  of  the  quadrilateral  the  equations  of 
whose  sides  are  x  =  0,  x+y=Q,  x+2y=5,  and  6x+y-\-58=0? 

4.  Obtain  the  formula  for  the  area  of  a  triangle  by  drop- 
ping perpendiculars  from  each  of  the  three  vertices  upon  the 
X-axis,  and  considering  the  trapezoids  formed. 

GENERAL  PROBLEMS 

1.  Show  that  the  triangle  whose  vertices  are  (3,  2), 
(— 1,   —3),  and  (—6,  1)  is  a  right  triangle. 

2.  An  isosceles  right  triangle  is  constructed  with  the  hy- 
potenuse on  the  line  x  -f  4  y  =  10,  and  the  vertex  of  the  right 
angle  at  the  point  (3,  4).  Find  the  coordinates  of  the  other 
vertices. 

3.  Find  the  equation  of  the  line  through  the  point  (5,  fi) 
which  forms  with  the  axes  a  triangle  whose  area  is  80.  Four 
solutions. 

4.  Find  the  equation  of  a  line  through  the  point  (—1,  5) 
such  that  the  given  point  bisects  that  portion  of  the  line  be- 
tween the  axes. 

5.  Find  the  equation  of  a  line  through  the  point  (3,  —  6) 
such  that  the  given  point  divides  that  portion  of  the  line  be- 
tween the  axes  in  the  ratio  3 :  —  1. 


Ch.  IV,  §40]  THE   STRAIGHT   LINE  61 

6.  Find  the  equation  of  the  line  passing  through  the 
point  (8,  2)  such  that  the  portion  of  it  included  between  the 
lines  x  —  2y  =  6  and  x  +  y  =  5  shall  be  bisected  at  the  given 
point. 

7.  On  the  line  y  —  5  =  0  a  segment  is  laid  off,  having 
for  the  abscissas  of  its  extremities  2  and  5,  and  upon  this 
segment  an  equilateral  triangle  is  constructed.  What  are  the 
coordinates  of  the  third  vertex? 

8.  Find  the  point  on  the  line  4?/  —  5x  +  28  =  0  which  is 
equidistant  from  the  points  (1,  5)  and  (7,  —  3). 

9.  Find  the  points  which  are  equidistant  from  the  points 
(4,  —  3)  and  (7,  1),  and  at  a  distance  3  from  the  line  15  x  -f-  8  y 
=  120. 

10.  The  coordinates  of  the  vertices  of  a  triangle  are  (5,  2), 
(4,  —  7),  and  (3,  7).  The  side  joining  the  first  two  points  is 
ilivided  in  the  ratio  4  :  7,  and  through  this  point  lines  are  drawn 
parallel  to  the  other  sides.  Find  their  points  of  intersection 
with  the  other  sides. 

11.  On  each  side  of  the  triangle  in  problem  10,  find  the 
point  which  is  equidistant  from  the  other  sides  of  the  triangle. 

12.  The  equations  of  the  sides  of  a  complete  quadrilateral 
are  2y  +  7  x=l4,  x—2y=l,  x+ky=—  4,  and  7 x— ky=—  28. 
Show  that  the  middle  points  of  the  three  diagonals  lie  on  a 
straight  line. 

13.  Show  that  the  perpendiculars  let  fall  from  any  point  of 
the  line  2  x  +  11  y  =  5  upon  the  two  lines  24  x  +  7  y  =  20  and 
4  x  —  3  y  =  2  are  equal  to  each  other. 

14.  Perpendiculars  are  dropped  from  the  point  (0,  4)  to  the 
sides  of  a  -triangle  whose  vertices  are  (1,  5),  (5,  —  1),  and 
(6,  0).    Show  that  the  feet  of  these  perpendiculars  lie  on  a  line. 

15.  Find  the  equation  of  a  line  through  the  intersection  of 
the  lines  2  x  —  7  y  =  3  and  x  +  3  y  =  8,  which  is  perpendicular 
to  the  line  joining  the  origin  to  the  intersection  of  these  lin»s 


62  ANALYTIC    GEOMETRY  [Ch.  IV,  §  40 

16.  Show  that  the  four  points  (2,  1),  (5,  4),  (4,  7),  and  (1,  4) 
are  the  vertices  of  a  parallelogram. 

17.  Find  the  area  of  the  triangle  formed  bylthe  three  lines 
y  =  mxx  +  (?!,  y  =  m<pc  -f-  c2,  and  x  =  0. 

18.  What  is  the  value  of  a  if  the  three  lines  3x  +  y  —  2  =  0, 
ax  +  2 y  —  3  =  0,  and  2x  —  y  —  3  =  0  meet  in  a  point  ? 

19.  Lines  are  drawn  through  the  vertices  of  a  triangle 
parallel  to  the  opposite  sides  of  the  triangle,  and  the  intersec- 
tions of  these  lines  are  joined  to  the  opposite  vertices  of  the 
triangle.     Show  that  the  joining  lines  meet  in  a  point. 

20.  Prove  analytically  that  the  bisector  of  the  interior  angle 
of  a  triangle  divides  the  opposite  side  into  segments  propor- 
tional to  the  adjacent  sides  of  the  triangle. 

21.  Prove  that  all  straight  lines,  for  which   -  +  -=-,  pass 

a      b      5 

through  a  fixed  point,  and  find  the  coordinates  of  that  point. 


CHAPTER  V 


POLAR   COORDINATES 


Fig.  39. 


41.  In  Art.  9,  the  polar  system  of  coordinates  was  men- 
tioned. Let  0  be  a  fixed  point  and  OA  a  fixed  line  through 
it.  Then  the  position  of  any  point 
P  is  fixed  if  the  angle  A  OP  and 
the  distance  OP  are  given.  The 
distance  OP  is  called  the  radius 
vector  of  the  point  P  and  is  rep- 
resented by  p.  Positive  values 
of  p  are  laid  off  from  0  along  the 

terminal  line  of  the  angle,  negative  values  in  the  opposite 
direction.  The  angle  A  OP  is  called  the  vectorial  angle 
and  is  represented  by  6.  The  usual  convention  in  regard 
to  angles  will  be  followed,  —  the  anti-clockwise  direction 

of    rotation   being    con- 

,p  sidered  positive.     These 

two  quantities  are  called 

the   polar   coordinates  of 

the  point,  and  are  written 

(/o,  0).     The  line  OA  is 

called  the  initial  line,  and 

the   point  0,    the   origin 

or  pole. 

It  appears  that,  while  any  pair  of  coordinates  determine 

a  single  point,  there  will  be  an  indefinite  number  of  pairs 

of  coordinates  which  will  give  the  same  point ;   for  there 

63 


64  ANALYTIC   GEOMETRY  [Ch.  V,  §  42 

will  be  an  indefinite  number  of  angles  which  have  the 
same  terminal  line.  If  6  is  restricted  to\values  between 
—  2  7r  and  2  7r,  any  point  may  be  determined  by  four  sets 
of  coordinates. 

If  the  polar  coordinates  of  the  point  P  in  Fig.  40 
are  (/?,  0),  the  same  point  may  also  be  determined  by 
(_p,  0  +  180°),  (-p,  (9-180°),  and  0,0-360°). 

PROBLEMS 

1.  Plot    the    following    points:    (%  j\  (6,  f  *),  ($,  -|\ 

(-10,  -f,),  (-2,1),  (4,0),  (-5,  0),  (0,tt). 

2.  Write  polar  coordinates  of  each  point  in  problem  1,  in 
which  p  and  0  are  both  positive. 

3.  Show  that  the  distance  between  two  points  whose  polar 
coordinates  are  (p1}  0X)  and  (p2,  02)  is  Vpi+pf— 2plp2cos(01— 62). 

42.  Equation  of  a  locus.  —  If  the  law  according  to  which 
a  point  moves  is  stated,  it  may  often  be  translated  into  an 
equation  connecting  p  and  0.  For  example,  if  a  point  is  to 
remain  at  a  distance  a  from  the  origin,  the  value  of  p  for 
every  such  point  is  a,  while  6  can  vary  at  pleasure.  The 
polar  coordinate  equation  of  a  circle  about  the  origin  is, 
therefore,  p  =  a.  The  equation  of  any  line  through  the 
origin  is  evidently  0  =  k,  for  on  such  lines  the  value  of  p 
is  entirely  unrestricted,  while  0  is  fixed. 

Again,  as  in  Cartesian  coordinates,  an  equation  connect- 
ing p  and  6  restricts  the  points  which  satisfy  it  to  a  series 
of  positions  which  lie  on  some  curve.  The  curve  which 
contains  all  the  points  whose  coordinates  satisfy  an  equa- 
tion and  no  other  points  is  called  the  locus  of  the  equation, 
and  the  equation  is  spoken  of  as  the  equation  of  the  locus. 


Ch.  V,  §  43]  POLAR  COORDINATES  65 

PROBLEMS 

1.  What  is  the  polar  coordinate  equation  of  a  line  which 

makes  an   angle  of  —  with  the  initial  line  and  which  passes 
4 

through  the  origin? 

2.  What  is  the  equation  of  aline  parallel  to  the  initial  line 
and  three  units  above  it  ? 

3.  Show  that  the  equation  of  any  line  in  terms  of  a  and  p  is 

p  cos  (0  —  «)  =  p. 

4.  Show  that  the  equation  of  a  circle  of  radius  r  about  the 
point  (pj,  Ox)  is       p2  +  p2  —  2  px  p  cos  (0  —  0X)  =  j-2. 

5.  A  circle  of  radius  r  has  its  centre  on  the  initial  line  and 
passes  through  the  origin ;  show  that  its  equation  is 

p  =  2  r  cos  6. 

43.  Plotting  in  polar  coordinates.  —  The  method  of 
finding  the  curve  which  is  the  locus  of  any  equation  in 
polar  coordinates  is  similar  to  that  employed  in  rectangu- 
lar coordinates.  Sometimes  the  law  of  formation  may  be 
determined  directly  from  the  equation,  but  it  is  usually 
necessary  to  find  various  points  on  the  curve  by  giving 
values  to  one  of  the  coordinates  and  finding  the  corre- 
sponding values  of  the  other ;  these  points  are  then 
plotted  and  a  smooth  curve  drawn  through  them. 
Coordinate  paper  may  be  made  by  drawing  circles  about 
the  origin  at  a  unit's  distance  from  each  other,  and  lines 
through  the  origin,  making  any  convenient  angle  with 
each  other.  On  this  the  position  of  the  points  may  be 
fixed  accurately  without  measurement. 

For  convenience  in  plotting  we  insert  a  table  of  the 
natural  values  of  the  trigonometric  functions  for  every  5° 
from  0°  to  90°. 


M 


ANALYTIC   GEOMETRY 


[Ch.  V,  §  44 


44.    Natural  values  of  the  sines,  cosines,  tangents,  and 
cotangents. 


0 

sin  0 

cos  0 

tan  0 

COt0 

0° 

.0000 

1.0000 

.0000 

90° 

5° 

.0872 

.9962 

.0875 

11.430 

85° 

10° 

.1736 

.9848 

.1763 

5.671 

80° 

15° 

.2588 

.9659 

.2679 

3.732 

75° 

20° 

.3420 

.9397 

.3640 

2.747 

70° 

25° 

.4226 

.9063 

.4663 

2.145 

65° 

30° 

.5000 

.8660 

.5774 

1.732 

60° 

35° 

.5736 

.8192 

.7002 

1.428 

55° 

40° 

.6428 

.7660 

.8391 

1.192 

50° 

45° 

.7071 

.7071 

1.0000 

1.000 

45° 

cos  0 

sin  0 

COt0 

tan0 

0 



PROBLEMS 

1.  Plot  the  locus  of  the  equation  p  sin  0  =  5. 

By  the  aid  of  the  table  of  sines  given  above,  the  following 
points  are  seen  to  lie  on  the  locus:  (29,  10°),  (14.6,  20a), 
(10,  30°),  (7.8,  40°),  (6.5,  50°),  (5.8,  60°),  (5.3,  70°),  (5.08,  80°), 
(5,  90°). 

Plotting  these  points  on  the  coordinate  paper,  the  locus 
appears  to  be  a  straight  line  MNy  parallel  to  the  initial  line 
and  five  units  above  it. 

2.  Plot  the  locus  of  the  equation  p-  =  100  cos  2  0. 

If  we  let  0  vary  from  0°  to  45°,  the  following  points  will  be 
found  to  lie  on  the  locus  :  (10,  0°),  (9.7, 10°),  (8.7,  20°),  (7,  30°), 
(4,  40°),  (0,  45°).  There  will  also  be  the  points  (-  9.7,  10°), 
etc.  The  values  of  p  will  be  imaginary  for  values  of  0  between 
45°  and  135°,  and  there  will  therefore  be  no  real  points  corre- 
sponding to  these  values  of  0.  If  we  let  6  vary  from  135°  to 
180°,  the  following  points  will  be  found  on  the  locus  :  (0, 135°), 
(4,  140°),   (7,  150°),   (8.7,  160°),   (9.7,  170°),   (10,  180°).     Also 


Ch.  V,  §  44] 


POLAB    COORDINATES 


67 


(_  4,  140°),  etc.  If  we  let  0  vary  from  180°  to  360°,  p  will 
pass  through  the  same  changes  in  value  as  before,  and  the 
same  points  will  be  located.     The  curve  has  the  form  shown 


Fig.  41. 

in  Fig.  41.  The  two  tangents  to  the  curve  at  the  origin 
make  angles  of  45°  and  135°  with  the  initial  line,  and  are 
therefore  perpendicular  to  each  other.  The  curve  is  called  the 
lemniscate. 

3.    Plot  the  locus  of  each  of  the  following  equations: 


(a)  p  sin  0  =  a. 

(b)  P(l-cos0)  =  2a. 

(c)  p  =  2a(l-cos0). 

(d)  P2  =  a2sin2  0. 

(e)  p  cos  0  =  a  cos  2  0. 
(/)  p  =  a  cos  3  6. 

(g)  p  =  a  sin  4  6. 

(li)  p2  cos  0  =  a2  sin  3  0. 


(0 

0) 

<*) 

(m) 

00 

(0) 


p2  =  cr  cos  3  0. 
p  =  a(sec0  +  tan  0). 
P  =  a(cos20  +  sm26). 
p  =  2  a  tan  0  •  sin  0. 
p  =  a(l+2cos0). 
p  =  aO. 
a 


P  = 


0 


V 


CHAPTER    VI 


TRANSFORMATION   OF   COORDINATES 


45.  When  the  equation  of  a  curve,  referred  to  any  sys- 
tem of  coordinates,  is  known,  it  is  often  desirable  to  obtain 
the  equation  of  the  same  curve,  referred  to  some  other 
system.  If  we  know  its  equation  in  Cartesian  coordinates, 
we  may  wish  to  obtain  its  equation  in  polar  coordinates, 
or  the  reverse.  Or,  knowing  its  Cartesian  equation  re- 
ferred to  a  certain  set  of  axes,  we  may  wish  to  obtain  its 
equation  referred  to  some  other  set  of  axes,  in  order  to 
obtain  the  simplest,  or  most  useful  form  of  its  equation. 
This  can  be  done,  if  we  can  obtain  the  relation  connecting 
the  coordinates  of  any  point  on  the  curve  in  the  first  sys- 
tem of  coordinates  and  the  coordinates  of  the  same  point 
in  the  second  system. 

We  can   transform  from  any  Cartesian  system  to  any 

other  by  first  changing 
the  origin  without  chang- 
ing the  direction  of  the 
axes,  and  then  revolving 
.,  each  of  the  axes  through 
some  angle. 

46.     Transformation  to 

axes  parallel  to  the  original 

axes.  — Let  OX  and  OY 

Fig.  42  be  any  given  pair  of  rec- 

68 


■X' 


M 


Ch.  VI,  §  40]    TRANSFORMATION   OF   COORDINATES  69 

tangular  axes,  and  let  0' X'  and  0' Y'  be  a  new  pair,  par- 
allel to  the  old,  and  having  for  their  origin  the  j)oint  Q'r 
whose  coordinates  with  respect  to  the  original  axes  are  x0 
and  y0.  Let  P  be  any  point,  and  let  its  coordinates  in 
the  first  system  be  (x,  y)  and  in  the  second  (V,  y'). 

From  the  figure,     031=  OA  +  AM, 
and  MP  =  AO'  +  NP. 

But  0M=  x,    OA  =  x0,   A3f=  x\ 

MP  =  y,   AO'  =  y0,    NP  =  y'. 

Substituting  these  values,  we  find 

«  =  «*  +  *, 
and  y  =  2/0  +  y'  L      J 

as  the  equations  connecting  the  old  and  new  coordinates, 
and  these  equations  will  be  found  to  hold  wherever  the 
point  P  is  placed  in  the  plane. 

If  we  have  an  equation,  which  expresses  the  law  of 
movement  of  a  point  by  giving  the  relation  between  its 
coordinates  referred  to  the  first  pair  of  axes,  the  substi- 
tution of  these  values  for  x  and  y  will  give  the  relation 
which  must  exist  between  x'  and  y1 ',  the  coordinates  of 
the  point  referred  to  the  second  pair  of  axes,  in  order 
that  the  point  may  move  in  the  same  path.  It  must  be 
understood  that  x'  and  y'  are  variables,  like  x  and  y. 
The  primes  are  only  used  to  distinguish  the  coordinates 
used  in  the  two  systems,  and  may  be  dropped  after  the 
substitution  has  been  made. 

In  the  above  demonstration  no  use  has  been  made  of  the 
fact  that  the  axes  are  rectangular.  The  same  formulas 
will  therefore  hold  for  transforming  from  any  set  of 
oblique  axes  to  any  parallel  set. 


TO 


ANALYTIC    GEOMETRY 


[Ch.  VI,  §  47 


PROBLEMS 

1.  If  the  equation  of  a  line  referred  to  any  given  system  of 
Cartesian  coordinates  is  3  x  +  4  y  =  10,  what  is  its  equation 
referred  to  a  parallel  system,  the  coordinates  of  whose  origin, 
referred  to  the  original  axes,  are  (—2,  5)  ? 

The  formulas  connecting  the  old  coordinates  of  any  point 
with  the  new  are 9   ,      , 

y  =  5  +  y'. 

Substituting  these  in  the  equation  of  the  line,  we  have 

3(-2-f*')  +  4(5  +  v/')  =  10, 

or  reducing  and  dropping  primes, 

3  x  -f  4  y  =  —  4. 

Construct  the  two  sets  of  axes  and  plot  the  locus  of  each  of 
the  equations,  showing  that  the  same  line  will  be  obtained  in 
both  cases. 

2.  The  equation  of  a  line  is  Ax  —  3y  =  S.  Find  the  equa- 
tion of  the  same  line,  referred  to  a  set  of  axes,  parallel  to  the 
old,  through  the  point  (2,  —  5)  as  origin. 

Plot  the  locus  with  respect  to  both  axes. 

47.  Transformation 
from  one  set  of  rectangu- 
lar axes  to  another,  hav- 
ing the  same  origin  and 
making  an  angle  8  with 
the  first  set.  —  Let  OX 
and  0  Y  be  the  given  set 
of  rectangular  axes,  and 
OX'  and  OY'  another 
set  of  rectangular  axes 
making  an  angle  6  with  the  first  set.  Let  the  coordinates 
of  P  with  respect  to  the  original  axes  be  (#,  y),  and  with 


Fig.  43. 


Ch.  VI,  §  48]    TRANSFORMATION   OF   COORDINATES  71 

respect  to  the  new  axes  (V,  //').  Draw  its  ordinates,  MP 
and  NP,  and  the  lines  KN  and  LN  parallel  to  OX  and 
OY.     The  angle  at  P  is  evidently  equal  to  6. 

OM=x,  MP  =  y,   ON=x\  NP  =  i/'. 

Then  0M=  OL-KN. 

But  OL  =  OxYeos  0  =  x'  cos  ft 

and  KN=  NP  sin  0  =  y'  sin  (9. 

Hence  as  =  a*'  cos  0  -  2/'  sin  0.  [21,  a] 

In  like  manner 

MP  =  LN  +  KP 

=  OiVsin<9  4-iVTPcos0 
or  y  =  a?'  sin  9  +  2/'  cos  0.  [21,  ft] 

48.  Transformation  in  which  both  the  position  of  the 
origin  and  the  direction  of  the  axes  are  changed.  —  If  it  be 
required  to  change  the  position  of  the  origin  to  the  point 
(x0,  y0),  and  at  the  same  time  to  revolve  the  axes  through 
the  angle  0,  the  two  operations  may  be  performed  sepa- 
rately, or  we  may  combine  the  two  previous  formulas  into 

the  one  set,  ,       .        ,  ,    M 

05  =  OCo  4-  oc'  cos  0-2/'  Sill  0, 

r221 

y  =  2/o  +  x'  sin  0  +  y'  cos  0.  *--"* J 

PROBLEMS 

1.  Transform  the  equation  3  x  +  7y  =  8  to  a  new  set  of 
axes  parallel  to  the  old  set,  and  having  the  point  (4,  —  2) 
as  origin. 

2.  Show  that  the  equation  x2  +  y2  =  a2,  referred  to  rectan- 
gular axes,  will  be  unchanged  by  revolving  the  axes  through 
any  angle,  keeping  the  origin  fixed. 

3.  Transform  the  equation  x2  —  y2  =  10,  referred  to  rectan 
gular  axes,  to  axes  bisecting  the  angle  between  the  old  axes. 


72 


ANALYTIC    GEOMETRY         [Cn.  VI,  §§  49,  50 


4.  Through  what  angle  must  the  coordinate  axes  be  turned, 
if  in  its  new  position  the  X-axis  goes  through  the  point  (5,  7)  ? 

5.  Given  the  equation  xr  +  y2  -4-  8  x  —  ky  =  0.  To-  what 
point  must  the  origin  be  changed  to  cause  the  terms  in  x  and  y 
to  disappear  ? 

6.  Given  the  equation  2y2  +  2  xy  +  r  +  4  =  0,  referred  to 
rectangular  axes.  Through  what  angle  must  the  axes  be 
turned  to  cause  the  term  in  xy  to  disappear  ? 


Fig.  44. 

% 

Let  the  student  show  that 


49.  Transformation  from 
any  Cartesian  system  to  any 
other  Cartesian  system,  hav- 
ing the  same  origin.  —  In 
Fig.  44,  OX  and  OY  are 
the  original  axes,  and  co  is 
the  angle  between  them  ; 
OX'  and  OY'  are  the  new 
axes;  OX'  and  OY'  make 
angles  6  and  cj>  with  OX. 


sin  w 


r/sinO-4>)3 
sinw 


sin  w 


y'Sin*. 


[23] 


sin  « 

What  do  these  formulas  becotne  when  co  —  90°  ?  When 
co  =  90°  and  $  =  6? 

50.  Degree  of  an  equation  not  changed  by  transformation 
of  coordinates.  —  The  degree  of  an  equation  cannot  be 
changed  by  transformation  from  one  system  of  Cartesian 
coordinates  to  any  other.  For  we  have  seen  that  in  each 
case  we  replace  x  and  y  by  expressions  of  the  first  degree 


Ch.  VI,  §51]    TRANSFORMATION   OF   COORDINATES 


73 


in  xf  and  y\  and  that  therefore  the  degree  of  the  equation 
cannot  be  raised.  Neither  can  it  be  lowered,  for  it  would 
then  be  necessary  to  raise  the  degree  in  transforming 
back  to  the  original  axes,  since  we  must  obtain  the  origi- 
nal equation. 


51.  Transformation  from  rectangular  to  polar  coordi- 
nates. —  Let  it  be  required  to  find  the  equations  of 
transformation  for  transforming  from  a  given  set  of  rec- 
tangular axes,  OX  and  OY,  to  a  polar  system  having  0 
as  its  origin  and  OX  as 
its  initial  line. 

The  relations  between 
^,  y,  />,  and  6  are  seen  at 
once   from   the   triangle 

OMP;    for   sin0  =  f^, 

,  n          OM  U±^        

and  cos  u  —  —— ,  or 


<JC  =  p  cos  e, 
y  =  psin  9. 


[24] 


M 


-X 


Fig.  45. 


The  formulas  for  transformation  from  polar  to  rectan- 
gular coordinates  are  easily  seen  from  the  same  triangle 
to  be 

,*,.  [25] 


8  =  tan 


It  is  not,  however,  generally  necessary  to  use  this 
second  set  of  formulas,  as  the  transformation  from  polar 
to  rectangular  coordinates  can  usually  be  made  more 
easily  by  the  aid  of  the  first  set. 


74  ANALYTIC   GEOMETRY  [Ch.  VI,  §  51 

PROBLEMS 

1.  Obtain  the  polar  equation  of  the  curve  whose  rectangular 
equation  is   x2  -f-  y2  =  r2. 

Substituting  x  =  p  cos  6,  and  y  =  p  sin  6,  we  have 

9  9/li  9*9/1  9 

p-  cos-  v  +  p-  sin-  0  =  r, 
or  p2  =  r2,  or  p  =  r. 

This  is  then  the  polar  equation  of  the  curve  whose  rectangu- 
lar equation  is  xr  +  y-  =  r2. 

2.  Obtain  the  polar  equations  of  the  curves  whose  rectangu- 
lar equations  are 

(a)  a2x2-\-b2y2  =  a2b2,  (e)  (x2  +  iff  =  ±a2x2y2, 

(b)  y2  =  2  mx,  (/)  a:2  +  f  +  **  =  0, 

(c)  x2-y2  =  a2,  (g)  y2  =  7>   *_    , 


(cf)   (ar  +  ?/2)2  =  o2  (.r3  —  ?/2),     (/t)  x2  +  y2  +  2  ax  =  aVx^  +  tf. 

3.  Obtain  the  rectangular  equation  of  the  curve  whose  polar 
equation  is  p  =  a  cos  0. 

We  might  make  this  transformation  by  using  the  two  for- 
mulas [25],  but  it  will  be  found  to  be  easier  first  to  multiply 
both  members  of  the  equation  by  p,  giving 
p-  =  ap  cos  6. 

Using  the  formulas  x  =  p  cos  0  and  p2  =  x2  +  y2,  this  reduces 
at  once  to  x2  +  y2  =  ax. 

This  is  then  the  rectangular  equation  of  the  curve  whose 
polar  equation  is  p  =  a  cos  0. 

4.  Obtain  the  rectangular  equations  of  the  curves  whose 
polar  equations  are 

(a)  p  =  a  sin  6,  (/)  p  —  a  sin  2  6, 

(b)  p  =  a  +  ——,  (g)  p2cos2  0  =  a2, 

(c)  p  =  a  (1  +  cos  0),  (li)  P  =  a  (cos  20  +  sin  2  (9), 

(d)  p2  =  a2  cos  2  6,  (i)   p  =  a  (1  +  cos  2  0), 

(e)  p  =  a  —  b  cos  0,  (,;')  p  =  2  a  tan  0  •  sin  0. 


CHAPTER    VII 


THE   CIRCLE 

52.  Equation.  —  The  locus  of  points  equidistant  from 
any  fixed  point  is  called  a  circle.  Hence,  to  find  the 
equation  of  a  circle,  it 
is  necessary  to  express 
the  algebraic  relation  be- 
tween the  coordinates  of 
such  points. 

If  the  origin  is  taken 
at  the  centre  of  the  circle, 
the  equation  is  evidently 


z2  + 


*2,    where 


Fig.  46. 


the  radius  of  the  circle. 
For  the  distance  of  any 
point  (#,  y)  from  the  origin  is  Vx2  +  y2. 

If  the  centre  be  taken  at  any  point  C,  whose  coordinates 
are  (a,  /3),  the  distance  CP  from  the  centre  to  any  variable 
point  P  is  w(x  —  a)2  +  (y  —  /3)2.  Hence  the  equation  of 
the  circle  is 


If  the  centre  is  on  the  -X"-axis,  /3  =  0, 

reduces  to 

(x  —  «)2  +  y2  =  r2; 


[26] 

and  the  equation 


if  on  the  P"-axis,  a  =  0,  and  the  equation  reduces  to 

x2  +  iy  -  /3)2  =  r2. 
75 


76  ANALYTIC    GEOMETRY  [Ch.  VII,  §  53 

Problem.  —  Find  the  equation  of  a  circle,  (a)  tangent  to 
both  axes  ;  (b)  passing  through  the  origin  and  having  its  centre 
on  the  X-axis. 

53.    General  form  of  the  equation.  —  Expanding  [26], 

we  have 

x2  +  yi  _  2  ax  -  2  /3y  +  a2  +  /32  -  r2  =  0. 

a  and  ft  can  have  any  value,  positive  or  negative,  and  r 
can  have  any  positive  value.  Hence  the  equation  is  in 
the  general  form  of 

#2  +  2/2  +  Zto  +  Ey  +  F=0,  [27] 

where  D  =  -2a,  U  =  -  2  j3,  and  F  =  a2  +  /S2  -  r2.  And 
if  an  equation  is  to  represent  a  circle,  it  must  be  in  the 
form  of  [27].  It  will  be  noted  that  this  is  not  the  most 
general  form  of  the  equation  of  the  second  degree.  For 
tins  IS 

Ax2  +  Bxy  +  C>2  +  Bx  +  Fy  +  F  =  0. 

When  the  two  equations  are  compared,  it  will  be  seen 
that  the  term  in  xy  is  wanting  in  [27],  and  that  the  coeffi- 
cients of  x2  and  y2  are  equal,  or 

B=0,  and  A  =  C. 

Hence  both  these  conditions  must  be  satisfied  in  order 
that  the  general  equation  of  the  second  degree  may  repre- 
sent a  circle. 

But  will  it  always  represent  a  circle  when  these  condi- 
tions are  satisfied  ?  It  will  be  necessary  to  determine 
whether  there  are  always  values  of  a,  /3,  and  r  which  cor- 
respond to  all  values  of  D,  F,  and  F.  Solving  the  equa- 
tions given  above  for  «,  /3,  and  r,  we  have 

-B        n         -  B 


/3  =  —^,    and   r  =  l^B2  +  F2-±F. 


Ch.  VII.  §  54]  THE   CIRCLK  77 

Hence  there  will  always  be  real  values  for  a  and  /3  for 
all  values  of  D,  K  and  F.  But  if  D2  +  F2  -  4  F <  0,  the 
value  of  r  is  imaginary,  and  there  will  be  no  point  in  the 
plane  which  will  satisfy  the  equation.  But  since  it  has 
the  form  of  the  equation  of  a  circle,  it  is  said  to  represent 
an  imaginary  circle. 

Again,  if  D2  +  F2  —  4  F=  0,  r  =  0,  and  the  equation 
represents  the  point  («,  /3)  onl}r.     It  is  called  a  null  circle. 

We  see  then  that  we  shall  have  a  real  circle  only  in 
case  D2  +  F2  —  4jP>  0.  But  no  equation  in  the  form 
of  [27]  can  represent  any  other  locus.  Hence  it  is  said 
to  represent  a  circle, 

real,  if  I)2  +  F2  -  4  F  >  0  ; 

null,  if  D2  +  F2-4F=Q  ; 

imaginary,  if  D2  +  F2  -  4  F  <  0. 

54.  Circle  through  three  points.  —  We  know  from  plane 
geometry  that  three  points  not  in  a  straight  line  determine 
a  circle.  It  ought  therefore  to  be  possible  to  find  the 
equation  of  the  circle  passing  through  three  such  points, 
(xv  3/j),  (x2,  #2),  and  03,  y3).  This  may  be  done  by 
determining  2),  F,  and  F  of  the  general  equation  [27] 
so  that  these  coordinates  will  satisfy  that  equation. 

Substituting  these  coordinates  successively  in  equation 
[27],  we  have 

x2  +  y2  +  DXl  +  Eyi  +  F  =  Q, 

x2  +  y2  +  Dx2  +  Fy2  +  F=0, 

From  these  three  equations  it  is  always  possible  to 
determine  D,  F,  and  F  (if  the  three  points  do  not  lie  on  a 


78  ANALYTIC    GEOMETRY  [Ch.  VII,  §  54 

line),  and  their  values  substituted  in  the  general  equation 

[27]    will  give   the   equation   of   the   circle   through   the 

points. 

PROBLEMS 

1.  What  is  the  equation  of  a  circle,  if 

(«)    its  centre  is  at  the  point  (—2,  3),  and  r  =  G, 

(b)  its  centre  is  at  the  point  (—  3,  —4),  and  r  =  5, 

(c)  its  centre  is  at  the  point  (5,  3),  and  it  is  tangent  to 
the  line  3  a,-  —  2y  =  10, 

(d)  its  radius  is  10,  and  it  is  tangent  to  the  line  4  x+oy  =  70 
at  the  point  (10,  10), 

(e)  it  passes  through  the  three  points  (4,  0),  (  —  2,  5),  (0,  —3), 
(/)  it  circumscribes   the  triangle,  the  equations   of   whose 

sides  are  x  -+-  2  y  —  5  =  0,  2  x  +  y  —  7  =  0,  and  x  —  y  -f  1  =  0, 
(r/)   it  has  the  line  joining  the  points  (3,  4)  and  (—  2,  0)  as  a 

diameter, 

(h)    it  passes  through  the  points  (5,  —  3)  and  (0,  G)  and  has 

its  centre  on  the  line  2  x  —  3  y  =  6, 

(i)    it  passes  through  the  points  (5,  —3)  and  (0,  6)  and  r=6? 

2.  Find  the  coordinates  of  the  centre  and  the  radius  of  each 
of  the  following  circles  : 

(a)  x2  +  y2  +  $x-6y-10  =  0, 

(6)  x2  +  y~  +  8  x  —  6  y  -f-  50  =  0, 

(c)  ar  +  ?/2 +  6?/-16  =  0, 

(d)  3ar  +  3?/2-7a:-8  =  0. 

3.  Show  that  if  the  equations  of  two  circles  differ  only  in 
the  constant  term,  they  represent  concentric  circles. 

4.  Show  that  the  equation  of  a  circle  in  oblique  coordinates 
is  in  the  form  of 

xr  +  2  cos  o)  •  xy  +  if  +  Dx  +  Ey  +  F=0. 

What  conditions  must  be  satisfied  by  the  general  equation 
of  the  second  degree  that  it  may  represent  a  circle  when  referred 
to  any  particular  set  of  oblique  coordinates  ? 


Ch.  VII,  §55]  THE   CIRCLE  7(.» 

5.  Show  that  the  equation  of  any  circle  through  the  points 
of  intersection  of  two  given  circles, 

*2  +  //'"'  +  Dp  4-  E&  +  F,  =  0, 

and  x2  -f  }r  +  D.2x  +  E,y  +  F,  =  0, 

can  be  expressed  in  the  form 

x-  +  if  +  Dxx  +  E&  +  F,  +  %  (.r  +  //2  +  XV?  +  Eojy  +  F2)  =  0. 

What  is  the  locus  of  this  equation  when  k  =  —  1  ? 

6.  Obtain  the  equation  of  the  common  chord  of  the  two 
circles, 

x-  +  y-  +  6x-y    =  0, 

and  ar  +  y-  —  4  y  +  10  =  0, 

and  show  that  it  is  perpendicular  to  their  line  of  centres. 

7.  Prove  that  the  common  chord  of  any  pair  of  intersecting 
circles  is  perpendicular  to  their  line  of  centres. 

8.  What  would  be  the  statement  of  problems  5  and  7,  if  the 
two  circles  do  not  intersect  ? 

55.  Tangent.  —  A  tangent  to  any  curve  is  denned  as 
follows :  Let  a  secant  through  a  fixed  point  P1  of  the 
curve  intersect  the  curve  again  at  P2.  Let  P2  move 
along  the  curve  toward  Pv  The  secant  will  revolve 
about  Pv  and  as  P2  approaches  Px  the  secant  will  ap- 
proach a  certain  limiting  position.  This  line,  which  is 
the  limiting  position  approached  by  the  secant  as  P2  ap- 
proaches Pv  is  called  the  tangent  to  the  curve  at  Pv 

The  method  of  finding  the  equation  of  the  tangent  to 
any  curve  of  the  second  degree  is  the  same  for  all.  The 
demonstration  should,  therefore,  be  studied  carefully  in 
the  case  of  the  circle  where  the  work  is  the  simplest. 


80 


ANALYTIC   GEOMETRY 


[Ch.  VII,  §  55 


According  to  the  definition  we  must  first  write  the 
equation  of  a  secant  through  two  points,  and  then  find 
the  limiting  form  which  this  equation  approaches  when 
the  two  points  approach  coincidence. 

Let  (xv  y{)  and  (x1  +  \  y^  +  &)  be  the  coordinates 
of  Px  and  P2,  adjacent  points  on  the  circle  x2  +  y2  =  r2. 
The   equation   of  the  line   through  these  two   points   is 

(by  [5]) 

y  —  y\  =  k. 

x  —  xx      h 

If  we  let  P2  approach  Pv  h  and  k  will  approach  zero, 
and  the  limit  of  the  second  member  will  be  indeterminate. 

This  would  be  neces- 
sary since  we  have 
made  no  use  of  the 
fact  that  P2  must  ap- 
proach Px  along  the 
circle.  Unless  P2  ap- 
proaches Px  along 
some  curve,  PXP2  will 
have  no  limiting  posi- 
tion. It  will  there- 
fore be  necessary  to 
determine    in    the  case  of   each  curve   the   value   of   the 

k 
expression  -■     In  the  case  of  the  circle  about  the  origin, 
ri 

the  coordinates  of  the  points  Px  and  P2  must  satisfy  the 

equation  x2  +  y2  =  r2. 

We  have,  therefore,       (1)  rrx2  +  y^  =  r2, 


Fig.  47. 


and        (2)  x2  +  2hx1  +  h2  +  y2  +  2  kyl  +  k2  =  r2. 


Ch.  VII,  §  50 J  THE    CIRCLE  81 

Subtracting  (1)  from  (2),  we  have 

2  hxx  +  A2  +  2  kyt  +  k2  =  0, 

k 
or,  transposing  and  solving  for  -, 

ill 

h_      2  xx  +  ht 

h~      2yi  +  k 

Substituting  in  the  former  equation  of  the  secant  PXPV 
we  see  that  y  _  ,fi  _       2x^  +  h 

x  —  xx  2y1-\-k 

is  another  form  of  its  equation  in  the  circle  x2  -f  y2  =  r2. 
If  now  we  let  h  and  k  decrease,  the  limit  of  the  second 

member  is   no   longer    indeterminate,   but  becomes K 

The  equation  of  the  tangent  is  therefore  * 

V-Vx=      \ 
x  -  xx         yl 

which  by  the  aid  of  (1)  reduces  to 

scias  +  y\V  =  r2.  [28] 

Let  the   student  show  by  the   same   method  that   the 

equation  of  the  tangent  to  the  circle 

a*  +  yZ  +  Dx  +  IJy  +  F=0 

is  xix  +  2/12/  +?(<*>  +  ^)  +  f  (V  +  2/i)+  F  =  0.  [29] 

56.  Normal.  —  The  normal  at  any  point  of  a  curve  is 
the  line  through  the  point,  perpendicular  to  the  tangent  at 
the  point.  Its  equation  can  be  obtained  by  first  writing 
the  equation  of  the  tangent  at  the  point,  and  then  that  of 
a  perpendicular  to  it  through  the  point  of  contact. 

The  equation  of  the  normal  to  the  circle  x2  +  y2  =  r2,  at 
the  point  (xxy^)^  is  seen  to  be  yxx  —  xxy  =  0. 


82  ANALYTIC   GEOMETRY  [Ch.  VII,  §  57 

PROBLEMS 

1.  Obtain  the  equations  of  the  tangents  and  normals  to  the 
following  circles,  and  show  that  in  each  case  the  normal  passes 
through  the  centre  of  the  circle : 

(a)  x2  +  f  =  25,  at  (3,  4), 

(6)  x>  -jly +  2*- 4y +  5  =  0,  at  (-  1,  2), 

(c)  x2  +  y2  — 14  x  —  4  ?/  —  5  =  0,  at  the  points  whose  abscissas 
are  10. 

(d)  x2  -f-  y2  —  6  a*  —  14  ?/  —  3  =  0,  at  the  points  whose  abscissas 
are  9. 

2.  Find  the  angle  in  which  the  two  circles  x2  +  ?/2  —  4  x  =  1 
and  a*2  +  y2  —  2  ?/  =  9  intersect. 

Note. — The  angle  between  two  curves  is  the  angle  between  their 
tangents  at  the  point  of  intersection. 

3.  Show  that  the  following  circles  cut  each  other  orthogo- 
nally (or  intersect  at  right  angles) : 

tf  +  y2-    8a?  +  4y+    7  =  0, 

x2  +  y2  -  10  x  -  6  y  +  21  =  0. 

4.  Show  that  the  length  of  the  tangent  from  the  point 
(xd  Vi)  to  the  circle  x2  +  y2  -\-  Dx  +  Ey  -+-  F  =  0  is 


VV  +  2/f  +  Bxx  +  Jg?y,  +  if. 

Note.  — Use  the  right  triangle  having  for  its  legs  the  tangent  and  the 
radius  to  the  point  of  contact.  The  length  of  the  hypotenuse  is  the  dis- 
tance from  the  point  (a?i,  y{)  to  the  centre  of  the  circle. 

5.  What  is  the  length  of  the  tangent  from  the  point 
(-  2,  6)  to  the  circle  x2  +  y2  +  2y  =  5? 

57.  Tangents  from  an  exterior  point.  —  The  equation  of 
the  tangent  which  we  have  obtained  can  be  applied  only 
when  we  know  the  coordinates  of  the  point  of  contact 


Ch.  VII,  §  57] 


THE   CIRCLE 


83 


(  xv  !/{).  There  are  other  conditions  which  will  determine 
the  tangent.  Consider  first  the  tangent  from  a  given 
exterior  point.  The  method  of  procedure  may  here  be 
best   shown  by  an   illustration. 

Let  it  be  required  to  find  the  equation  of  a  tangent 
from  the  point  (5,  10)  to  the  circle  whose  equation  is 
.^  +  i/2  =  1oo. 

Let  the  coordinates  of  the  unknown  point  of  contact 
be  (xv  y^).  Then  the  equation  of  the  tangent  will  be 
xxx  +  yxy  =  100.  Now  this  tangent  is  to  pass  through 
the  point  (5,  10),  and  therefore  these  coordinates  must 
satisfy  its  equation,  or 

5^  +  10^  =  100. 

This  is  one  equation  connecting  x1  and  yv  and  the  fact 
that  the  point  (xv  yx}  lies  on  the  circle  gives  another, 

^2  +  ^2  =  100. 

The  algebraic  solution  of  these  equations  gives 


2-1  =  0, 
10, 


or 


^=10,         ^  =  6. 

There  are,  therefore,  as  we  should  expect,  two  points  of 
contact  of  tangents  from  the  given  exterior  point,  viz. :  (0, 10) 
and  (8,  6).  Substituting  these 
values  in  the  equation  of  the 
tangent,   we  have 


and 


10  #  =  100, 
8  x  +  6  y  =  100, 


as  the  equations  of  the  tangents 
through  the  point  (5,  10). 


84  ANALYTIC    GEOMETRY  [Ch.  VII,  §  58 

PROBLEMS 

1.  Obtain  the  equations  of  the  tangents  to  the  following 
circles : 

(a)  x-  +  y~  =  49,  from  (6,  8). 

(b)  x2  +  y2-±x-22  =  0,  from  (-  2,  6). 

(c)  x2  +  f  +  5  y  =  25,  from  (7,  -  1). 

2.  Obtain  in  each  of  the  problems  the  equation  of  the  line 
joining  the  points  of  contact  of  the  two  tangents. 

3.  Obtain  in  this  way  the  equation  of  the  chord  of  contact 
of  tangents  from  the  exterior  point  (xx,  y{)  to  the  circle 
x2  +  y2  =  r2. 

58.  Tangent  in  terms  of  its  slope.  —  When  the  slope  of 
the  tangent  is  given,  we  might  proceed  as  in  Art.  57,  for 
we  could  obtain  one  equation  by  placing  the  slope  of  the 

X-i 

tangent, -,  equal   to   the   given  slope.      Solving  this 

with  xf  +  y^  =  100,  we  could  find  xx  and  yx  just  as  before. 

But  another  method  is  more  important.     The  equation  of 

any  line  which  has  the  given  slope  I  may  be  written  in  the 

form  7     ,   , 

y  =  lx  +  b. 

It  is  then  only  necessary  to  find  what  value  of  b  will 
make  it  a  tangent  to  the  circle  x2  +  y2=  r2.  Every  line 
of  the  system  will  cut  the  circle  in  two  points,  real, 
imaginary,  or  coincident.  If  the  points  are  coincident, 
the  line  is  a  tangent.  Starting  the  solution  of  y  =  lx  +  b 
and  x2  +  y2  =  r2,  we  have  at  once  by  substitution 

(l+/2)aa  +  2  lbx  +  b2- 7^  =  0 

for   determining   the    abscissas    of    the    points    of    inter- 
section. 


tin.  VII,  §  59]  THE   CIRCLE  85 

This  Avill  in  general  have  two  distinct  roots,  but  (by 
Art'8)if  (2  lb)*  =4(1  +  P)(J2-r»), 

these  roots  are  equal.  This  equation  therefore  gives  the 
value  of  £,  which  makes  the  line  a  tangent.  Solving  for 
b,  we  have 

b  =  ±  rvr+?. 

There  are  then  two  tangents  to  the  circle  which  have 
any  given  slope.      Their  equations  are 

y=lx±  rVTTW.  [30] 


PROBLEMS 

1.  Obtain  the  equations  of  the  tangents  to  the  circle 
x2  +  y2  =  49,  which  are  (a)  parallel  to  the  line  3  x  —  2  y  =  10 ; 
(b)  perpendicular  to  the  same  line. 

2.  Obtain  the  equations  of  the  tangents  to  the  circle 
x2  -|-  if  -f.  6  x  =  0,  which  are  perpendicular  to  the  line 
x-3y.+  4  =  0. 

3.  Determine  the  relation  between  a,  b,  and  r  if  the  line 

- '  +  --  =  1  is  tangent  to  the  circle  ar  +  y2  =  r2. 
a     b 

4.  Determine  the  value  of  k  if  the  line  3x  —  Ay  —  k  is 
tangent  to  the  circle  x2  -+-  y2  —  8  x  +  12  y  —  44  =  0. 

5.  Find  the  condition  which  must  be  satisfied  if  the  line 
Ax  +  By  +  C  =  0  is  tangent  to  the  circle 

x*  +  y2  +  Dx  +  Ey  +  F=0. 

59.  Chord  of  contact.  —  We  have  seen  that,  from  any 
point  Px  outside  the  circle  x2  +  y'z  =  r2,  two  tangents  can 
be  drawn  to  the  circle.     Let  it  be,  required  to  find  the 


ANALYTIC    GEOMETRY 


[Ch.  VII,  §  59 


equation  of   the  chord  P2P3  through  the  two   points  of 
contact  of  these  tangents. 

The   equations  of   the    tangents   P2PX    and   P3Px    are 

(by  [28]) 

and    x3x  +  y$  =  r2. 

Both   these   equations 
X    must  be  satisfied  by 

On  Vi)- 
Hence  x2xx  -f  y2yx  =  r2, 

and  x3xt  +  y3y1  =  r2. 

But  these  are  just  the 
conditions  which  must  be 
satisfied  if  the  points  P2  and  P3  are  on  the  line 

ocioc  +  y\\f  =  r2.  [31] 

This  is,  therefore,  the  equation  of  the  line  P2P3  which 
is  the  chord  of  contact. 

It  Avill  be  noted  that  this  equation  has  the  same  form  as 
the  equation  of  the  tangent.  It  represents  the  tangent  if 
the  point  Px  is  on  the  circle ;  but  if  Px  is  outside  the  circle, 
it  is  the  equation  of  the  chord  of  contact. 

Let  the  student  show  that  the  equation  of  the  chord  of 
contact  of  tangents  from  an  exterior  point  to  the  circle 


is 


0. 


[32] 


PROBLEMS 

1.    Find  the  length  of  the  chord  of  contact  of  tangents  from 
the  point  (3,  4)  to  the  circle  $  +  y2  =  4. 


Oh.  VII,  §69]  THE   CIRCLE  87 

2.  Find  the  equation  of  the  circle  which  touches  the  line 
2x  —  ?/ =  10  at  the  point  (3,  —4)  and  passes  through  the 
point  (5,  1). 

3.  Find  the  equation  of  the  circle  which  passes  through  the 
point  (1,  1)  and  also  through  the  intersections  of  the  circles 
'<•'  +  .'/"  —  •>  a;+4y=10,  and  x2+y2=5x.     [See  prob.  5,  page  79.] 

4.  Find  the  equations  of  the  three  common  chords  of  the 
three  circles  in  problem  1,  page  84,  and  show  that  they  inter- 
sect in  a  point. 

5.  Find  the  equation  of  the  circle  inscribed  in  the  triangle 
whose  sides  are  represented  by  the  equations 

4  x  -f-  3  y  —  10,  x  —  5  y  =  15,  and  3  x  —  4  y  =  8. 

6.  Find  the  area  of  the  triangle  formed  by  the  axes  of 
coordinates  and  the  tangent  to  the  circle  x2  +  y2  =  r2  at  the 
point  (i\,  yx). 

7.  Construct  the  circles  x2  +  y-  —  x  -+-  2  y,  and  x2  +  y2  =  2x. 
Find  the  equations  of  their  line  of  centres,  their  common  chord, 
and  points  of  intersection.  Show  that  their  common  chord  is 
perpendicular  to  their  line  of  centres.  At  what  angles  do  the 
circles  intersect  ? 

8.  Show  that  in  any  circle  a  line  perpendicular  to  the  tan- 
gent at  the  point  of  contact  passes  through  the  centre. 

9.  Show  that  an  angle  inscribed  in  a  semicircle  is  a  right 
angle. 

10.  Show  that  the  perpendicular  from  any  point  of  a  circle 
on  a  chord  is  a  mean  proportional  between  the  perpendiculars 
from  the  same  point  on  the  tangents  at  the  extremities  of  the 
chord. 

11.  Show  that  the  chord  of  contact  of  tangents  from  an 
exterior  point  is  perpendicular  to  the  line  joining  that  point 
to  the  centre  of  the  circle. 


CHAPTER   VIII 

LOCI 

60.  We  have  seen  that  when  a  property  common  to  all 
points  of  a  locus  is  given,  the  translation  of  this  property 
into  an  algebraic  equation  between  the  coordinates  of  the 
points  gives  the  equation  of  the  locus  ;  for  this  is  just  what 
is  meant  by  the  equation  of  a  locus,  —  an  equation  which  is 
satisfied  by  the  coordinates  of  every  point  which  satisfies 
the  given  conditions,  and  by  no  other  points.  The  actual 
Avork  then  always  consists  in  this  translation  of  a  condition 
expressed  in  language  into  a  relation  between  the  coordinates 
expressed  in  an  algebraic  equation.  Any  method  which 
enables  us  to  do  this  may  be  employed.  The  simple 
methods  have  already  been  exemplified  in  the  previous 
chapters.  In  these  cases  the  law  may  be  expressed  as 
an  equation  in  x  and  y  at  once  by  the  aid  only  of  a  sim- 
ple geometrical  construction.  There  are  many  problems 
which  may  be  solved  in  this  way. 

PROBLEMS 
1 .    The  sum  of  the  squares  of  the  distances  of  a  moving 
point  from  two  fixed  points  is  constant.     Find  the  locus 
of  the  moving  point. 

Let  the  .X-axis  pass  through  the  fixed  points,  with  the 
origin  midway  between  them.  Then  («,  0)  and  (—  a,  0) 
will  represent  the  points.     Let  (x,  y)  be  any  position  of 

88 


Ch.  VIII,  §  60] 


LOCI 


8y 


tlic  moving  point.     Then  placing  the  sum  of  the  squares 
of  the  two  distances  equal  to  a  constant,  Jc,  we  have 

+  [(.*  + a)2 +  </*]  =  &. 

This  is  then  the  equation  which 
must  be  satisfied  by  the  coordi- 
nates of  all  points  fulfilling  the 
given  conditions,  and  is  there- 
fore the  equation  of  the  locus 
desired.     Reducing,  we  have 


A" 


B(-a,o) 


A  (a,  o) 


?  +  y*  =  v. 


Yr 
Fig.  50. 


This  is  the  equation  of  a  circle  about  the  origin.  This 
property  of  a  circle  might  be  used  to  define  it  as  well  as 
the  more  familiar  one.  Most  curves  may  be  defined  in 
many  ways ;  for  any  property  which  is  sufficient  to 
determine  the  curve  completely  may  be  used  as  its 
definition. 

2.  The  difference  of  the  squares  of  the  distances  of  a 
moving  point  from  two  fixed  points  is  constant.  Show 
that  its  locus  is  a  line  perpendicular  to  the  line  through 
the  two  fixed  points. 

3.  The  distances  of  a  moving  point  from  two  fixed 
points  are  equal.     Find  the  locus. 

4.  The  distances  of  a  moving  point  from  two  fixed 
points  are  in  the  ratio  of  m  to  n.  Show  that  the  locus  is 
a  circle.  Find  the  centre  and  radius,  and  show  that,  if 
m  =  w,  the  locus  becomes  the  same  as  that  obtained  in 
the  previous  problem. 


90  ANALYTIC    GEOMETRY  [Ch.  VIII,  §  6ti 

5.  Find  the  locus  of  the  centres  of  circles  of  radius  r, 
which  pass  through  a  fixed  point  (xv  yx). 

6.  Find  the  locus  of  the  centres  of  circles  which  pass 
through  two  fixed  points. 

7.  Find  the  locus  of  the  centres  of  circles  which  touch 
two  given  lines. 

8.  The  sum  of  the  squares  of  the  distances  of  a  moving 
point  from  the  sides  of  a  square  is  constant.  Find  the 
locus. 

9.  Find  the  locus  of  a  point,  the  square  of  whose  dis- 
tance from  a  fixed  point  is  m  times  its  distance  from  a 
fixed  line. 

Note.  —  Take  the  fixed  line  as  the  Y-axis,  and  let  the  X-axis  pass 
through  the  fixed  point. 

10.  The  sum  of  the  squares  of  the  distances  of  a  moving 
point  from  the  four  corners  of  a  fixed  square  is  constant. 
Show  that  the  locus  is  a  circle  whose  centre  is  at  the  centre 
of  the  square. 

11.  Given  the  base  of  a  triangle  and  the  distance  from 
one  end  of  the  base  to  the  middle  point  of  the  opposite 
side,  find  the  locus  of  the  vertex. 

12.  The  sum  of  the  squares  of  the  perpendiculars  let 
fall  from  a  moving  point  on  the  sides  of  an  equilateral 
triangle  is  constant.     Find  its  locus. 

13.  The  sum  of  the  squares  of  the  distances  of  a  mov- 
ing point  from  r  fixed  points  is  constant.  Show  that  its 
locus  is  a  circle. 

14.  A  line  of  given  length  moves  so  that  its  ends  shall 
always  touch  two  lines  at  right  angles  to  each  other 
Find  the  locus  of  the  middle  point. 


Ch.  VIII,  §61]  LOCI  01 

15.  The  three  points  (9,  ilf,  iV*  lie  on  a  line.  Find  the 
locus  of  the  point  P,  when  ZOPM=ZMPN. 

16.  One  side  of  a  triangle  and  the  angle  opposite  are 
fixed.     Find  the  locus  of  the  vertex  of  the  angle. 

61.  It  is,  however,  often  impossible  to  obtain  easily 
by  direct  geometrical  methods  the  relation  between  the 
coordinates.  We  may  then  find  it  necessary  to  introduce 
certain  other  auxiliary  variables,  which  we  call  param- 
eters. These  must  be  so  chosen  that  it  is  possible  to 
express  in  equations  the  relation  between  the  coordinates, 
x  and  y,  of  the  moving  point  and  these  parameters.  If 
we  have  introduced  n  parameters  and  can  find  n  +  1 
independent  equations,  it  is  always  possible  to  combine 
them  in  such  a  way  as  to  eliminate  all  the  n  parameters 
and  leave  a  single  equation  connecting  the  coordinates  of 
the  moving  point.  This  resulting  equation  must  be  the 
equation  of  the  locus.  The  difficulty  of  the  elimination 
evidently  increases  with  the  number  of  parameters  used. 
When  more  than  two  or  three  are  used,  it  becomes  very 
laborious.  Care  should  therefore  be  taken  to  choose  that 
method  which  requires  the  introduction  of  the  fewest 
parameters. 

The  following  problems  illustrate  some  of  the  methods. 

17.  A  straight  line  is  drawn  parallel  to  the  base  of 
a  triangle  and  its  extremities  are  joined  transversely  to 
those  of  the  base.  Find  the  locus  of  the  intersection 
of  the  joining  lines. 

Choose  the  base  of  the  triangle  as  the  JT-axis  and  a 
perpendicular  to  this   side   through    the   opposite   vertex 


92  ANALYTIC   GEOMETRY  [Ch.  VIII,  §  61 

as  the  Y-axis.     Let  DE  be  any  position  of  the  line  par- 
allel to  the  base.     We  are  to  find  the  locus  of  j?;,  the 

point  of  intersection  of  the  lines 
b  (o,b)  CE  and  AD.     Let  the  coordinates 

of  A  be  (a,  0);  of  B,   (0,  b);    of 
x    C,    O,   0);    and  of  P',    (V,   /). 
A  (a,0)    Any  particular  values  of  x'  and  y 
evidently  depend  upon  the  posi- 
Y  tion  of  DE,  and  this  depends  upon 

a  single  parameter,  its  distance 
from  OA.  Let  the  equation  of  DE  in  any  particular 
position  be  y  =  k.  We  need  the  equations  of  the  lines 
CE  and  AD.  It  is  therefore  necessary  to  determine 
the  coordinates  of  D  and  E.     The  equation  of  AB   is 

-  4-  ^  =  1,  and   by  solving  this  equation  with   the  equa- 
a      b 

tion  y  =  &,  the  coordinates  of  i?  are  easily  found  to  be 
,  k).     In  like  manner  the  coordinates  of  D  are 


b 
found  to  be  f  -^ — - — s  k  J. 

The  equations  of  J.D  and  CE  are 

&&e  +  [(#  —  e)^  +  6'^]#  —  kab, 

and  &£>#  +  [(c  —  a)6  +  «&]#  =  kcb. 

Now  since  these  two  lines  both   pass  through  P\  its 
coordinates  (V,  y' )  must  satisfy  both  equations,  or 

kbx'  +  [(«  —  c)6  +  e&],/  =  &a&, 

and  kbx'  4-  [0?  —  &)&  +  ak~\y'  =  kcb. 

Here  then  are  two  equations  between  x\  y\  and   k. 
The  elimination  of  k  will  give  a  single   equation  in  x' 


Ch.  VIII,  §  01]  LOCI  93 

and  y'  which  must  be  the  equation  of  the  locus  of  P'. 
For  it  will  be  the  algebraic  expression  of  the  relation 
which  must  exist  between  the  coordinates  of  P\  that 
it  may  be  'the  intersection  of  the  two  diagonals.  The 
elimination  is  here  easily  performed.       For,  adding,  we 

have 

2  kbxr  +  k(a  +  e)y'  =  kb(a  +  c). 

Dividing  by  k  and  dropping  the  primes,  we  have  as 
the  equation  of  the  locus, 

2  bx  +  (a  +  c*)y  =  b(a  +  c'). 

Let  the  student  find  from  the  conditions  of  the  problem 
two  points  through  which  the  curve  must  pass,  and  test 
the  result  obtained  above  by  substituting  in  it  the  coor- 
dinates of  these  points. 

18.  Find  the  locus  of  the  intersection  of  the  diagonals 
of  rectangles  inscribed  in  a  given  triangle. 

19.  On  the  sides  of  a  given  triangle  measure  off  equal 
distances  from  the  extremities  of  the  base,  and  at  these 
points  erect  perpendiculars  to  the  sides.  Find  the  locus 
of  the  point  of  intersection  of  these  perpendiculars. 

20.  The  ends  of  the  hypotenuse  of  a  given  right  tri- 
angle touch  the  coordinate  axes.  Find  the  locus  of  the 
vertex  of  the  right  angle. 

21.  Parallel  lines  are  drawn  with  their  ends  on  the 
two  axes.  Find  the  locus  of  the  point  which  divides 
them  in  the  ratio  of  m  :  n. 

22.  One  side  and  the  opposite  angle  of  a  triangle  are 
fixed.  Find  the  locus  of  the  centre  of  the  inscribed 
circle. 


94  ANALYTIC   GEOMETRY  [Ch.  VIII,  §  61 

23.  Each  radius  of  the  circle,  x2  -f  y2  =  r2,  is  extended 
a  distance  equal  to  the  ordinate  of  its  extremity.  Find 
the  locus  of  its  terminal  point. 

24.  In  a  rectangle,  ABCD,  let  EF  and  GcH  be  drawn 
parallel  respectively  to  AB  and  BO.  Find  the  locus  of 
the  intersection  of  HF  and  EG-. 

25.  In  the  previous  problem  let  ABCD  be  any  paral- 
lelogram and  solve  with  the  aid  of  oblique  coordinates. 

26.  Find  the  locus  of  the  middle  point  of  a  system  of 
parallel  chords  of  the  circle  x2  -f  y2  —  r2. 

Let  y  =  Ix  +  b  be  the  equation  of  any  one  of  the  parallel 
chords ;  let  (xv  y^)  and  (z2,  y2)  be  the  coordinates  of 
the  points  where  it  cuts  the  circle,  and  (x\  y')  the  coor- 
dinates of  the  point  midway  between  these  points.  It 
is  required  to  find  an  equation  connecting  x'  and  y'  which 
may  contain  I  but  must  not  contain  b. 

Starting  the  solution  of  the  two  equations,  y  =  lx  +  b 
and  x2  -h  y2  =  r2,  we  have 

(1  +  l2)x2  +  2lbx  +  b2-r2  =  0, 

the  two  roots  of  which  must  be  x1  and  x2. 

But  x'  =  ^±^2- 


Hence  (1)  x'  =  -  -^~-    (See  Art.  8.) 

Since  the  point  (V,  y'}  lies  on  the  line  y  =  Ix  +  b,  its 
•coordinates  must  satisfy  that  equation,  or 

(2)  yf  =  Ix'  +  b. 


Ch.  VIII,  § 


LOCI 


95 


We  have  then  two  equations  in  a/,  y\  Z,  and  Z>,  from 
which  b  must  be  eliminated.  From  (2)  b  =  y' —  lx'.  Sub- 
stituting this  value  in  (1)  and 
reducing,  we  have  as  the  equa- 
tion of  the  desired  locus 

x  -f  ly  =  0. 

Since  this  equation  is  of  the 
first  degree  and  contains  no 
constant  term,  it  represents 
a  straight  line  through  the 
centre  of  the  circle,  and  con- 
forms to  the  ordinary  definition  of  a  diameter, 
evidently  perpendicular  to  the  parallel  chords. 


It   is 


27.  Find  the  Iogus  of  the  middle  points  of  chords  which 
pass  through  a  fixed  point  (xv  yx)  of  the  circle  x2  +  y2  =  r2. 

Let  P\  (V,  ?/ ),  be  the  middle  point  of  any  chord  through 
Pv  (xv  y{).  Let  (xvy^)  be  the  coordinates  of  P2,  the 
other   extremity  of  the  chord.     From    the    formulas    for 

bisecting  a  line  [4],  we  have 


(1)   x'  = 


x,  4-  x, 


and       (2)  y'  =  U±±l2. 

And,  since  P2  is  a  point  on 
the  circle, 


(3) 


+ 


Here  are  three  equations 
between  the  variables  xf  and  y\  the  constants  xv  yv  and  r, 
and  the  parameters  x2  and  yv     It  is  therefore  possible 


!»»; 


ANALYTIC    GEOMETRY 


[Ch.  VIII,  §  61 


to  eliminate  the  parameters  and  obtain  a  single  equation 
in  terms  of  the  variables  and  constants  only.  Solving  (1) 
and  (2)  for  x2  and  yv  we  have 

x2  =  2  x'  -  xv  and  y2  =  2  yf  -,  yv 
Substituting  these  values  in  (3),  we  have 


4  a/»  +  4  y'2  -  4  a^'  -  4  yty'  +  zx2  +  yx2  =  r2. 

But   a^2  +  j/j2  =  r2,  and,  dropping  primes,  the  equation 

reduces  to 

x2-{-y2-x1x-y1y  =  0. 


This  is  the  equation  of  the  locus  of  P' .     It  is  a  circle  on 

0P1  as  a  diameter,  since  its  centre  is  at  the  point  I  -J,  2l  J, 
and  it  passes  through  the  origin. 

When,  as  in  the  above  problem,  we  have  to  determine 
the  locus  of  a  point  situated  on  a  moving  line  which 
revolves  about  some  fixed  point  in  it,  polar  coordinates 
are  often  convenient.  The  fixed  point  is  taken  as  the 
pole,   and   the   distance   from   it  to   any  position    of   the 

Y^ moving     point     becomes 

"A   the  radius  vector. 

The  following  prob- 
lem will  illustrate  the 
method : 

28.  Find  the  locus  of 
the  middle  points  of 
chords  of  the  circle, 


;r 


r\ 


which  pass  through  a  fixed  point,   (xv   t^),   not   on   the 
circle. 


Ch.  VIII,  §61]  LOCI  97 

Le1  J\  be  a  fixed  point  through  which  the  secant  PXPS 
passes,  and  let  it  be  required  to  find  the  locus  of  _P',  the 
middle  point  of  P2Ps-  Transform  the  equation  of  the 
circle  to  polar  coordinates,  with  P1  as  origin.  The  equa- 
tions of  transformation  are  (by  [20]  and  [24]), 

x  =  X-,  4-  p  cos  6, 
(1)  1 

y  =  Vi  +  p  sin  0, 

and  the  equation  of  the  circle  becomes 

(2)  p2  +  2  (xx  cos  6  +  yx  sin  6)  p  +  xx2  +  y2  -  r2  =  0. 

Let  p'  and  6'  be  the  polar  coordinates  of  P' .  The  vec- 
torial angles  of  P2,  P' ',  and  P3  are  evidently  the  same, 
and  if  6'  be  substituted  for  6  in  (2),  the  solution  of  the 
resulting  equation, 

p2  +  2  Oj  cos  0'  +  y.x  sin  6")  p  +  ^2  +  yx2  -  r2  =  0, 

for  ^  will  give  p2  and  p3,  the  two  values  of  p  for  the  points 
P2  and  P3. 

But  /  =  ^, 

and         p.,  +  p3  =  —  2  (a^  cos  0r  +  ^  sin  0').      (See  Art.  8.) 

ence         p'  =  —  (xx  cos  6'  +  y1  sin0'). 

This  equation  expresses  the  relation  which  must  exist 

between  the  polar  coordinates  of  P' ',  and,  dropping  primes, 

we  have  as  the  polar  equation  of  the  locus,  referred  to  Px 

as  origin, 

p  =  —  xx  cos  6  —  yx  sin  6. 

The  equations  for  transforming  back  to  rectangular 
coordinates,  obtained  from  (1),  are 

p  cos  6  =  x  —  xv 
and  p  sin  6  =  y  —  yr 


98  ANALYTIC    GEOMETRY  [Ch.  VIII,  §  61 

From  these  we  see  that 

If  the  polar  equation  of  the  loeus  is  multiplied  by  p, 
and  these  values  substituted,  it  becomes 

(x  -  xxy  +  (y-  2/j)2  =  -  xxx  +  x2  -  yxy  +  y2, 

or  x2  +  y2  —  xxx  —  yxy  =  0. 

This  is  the  rectangular  equation  of  the  locus  referred  to 
the  original  origin,  and  is  seen  to  represent  a  circle  on 
OP1  as  diameter. 

29.  Solve  problem  27  by  means  of  polar  coordinates, 
and  problem  28  by  means  of  rectangular  coordinates. 

30.  Find  the  locus  of  the  points  which  divide  in  the 
ratio  m  :  n  chords  through  a  fixed  point  (xv  y-^)  of  the 
circle  x2  +  y2  =  r2. 

31.  Lines  through  a  fixed  point  P1  cut  the  circle 
x2  +  y2  =  r2  in  the  points  P2  and  Pg.  Find  the  locus  of  a 
point  P  of  this  line,  if 

P    p  =  ^  -Pl-**2   X     ™1™3, 

1    "  PlPi  +  PlPi 

32.  Chords  through  a  fixed  point  of  a  circle  are  extended 
their  own  length.     Find  the  locus  of  their  extremity. 

33.  Lines  are  drawn  from  a  fixed  point  Pv  meeting  a 
fixed  circle  in  P2.  On  PXP2  a  point  P  is  taken  so  that 
PXP  x  PXP2  =  k2.     Find  the  locus  of  P. 

34.  Lines  are  drawn  from  a  fixed  point  Pv  meeting 
a  fixed  line  in  P2.  Find  the  locus  of  the  point  which 
divides  PXP2  in  the  ratio  m  :  n. 


Cn.  vin,  §oi]  loci  99 

35.  Lines  are  drawn  from  a  fixed  point  Pv  meeting  a 
fixed  line  in  Pv  Find  the  locus  of  a  point  P  on  these 
lines  if  PXP  x  PYP2  =  P. 

36.  Find  the  locus  of  points  from  which  tangents  to 
two  given  fixed  circles  are  equal.  (See  problem  4,  page 
82.)  Show  that  the  locus  is  a  line  perpendicular  to  the 
line  joining  the  centres  of  the  two  circles. 


CHAPTER   IX 


M 


CONIC   SECTIONS 

62.    Definition  and  equation. — If  a  point  moves  so  that 
the  numerical  ratio  of  its  distance  from  a  fixed  point  to  its 

distance  from  a  fixed  line 
remains  constant^  its  locus 
is  called  a  conic. 

Let   the   fixed  line  be 
taken  as  the  Z"-axis,  and 

X    a  perpendicular  through 

the  fixed  point  F  as  the 
X-axis.     Let  the  perpen- 
dicular  distance    OF   of 
the  fixed  point  from  the 
fixed  line  be  represented  by  m.     Let  P  be  any  position 
of  the  moving  point.     Then  Ave  are  to  find  the  equation 
of  the  locus  of  P  when 
FP 
MP 


F 


Fig.  55. 


G) 


(any  constant)  =  e. 


But 


FP  =  V(x  -  m)2+  y\  and  MP  ==  x. 


Then  (1)  becomes     — ^ )   ~r  V  _  e<f 


or 
or 


x2  —  2  mx  +  m2  -f-  y2  —  e2x2, 

(1  -  e-)x2  -  2  mx  +  y-  +  m-  =  0. 
100 


[33] 


Ch.  IX,  §63]  CONIC   SECTIONS  101 

This  is  then  the  general  equation  of  a/ conic.  The  form 
which  it  takes  in  any  particular  case  depends  upon  the 
values  given  to  the  constants,  m  and  e. 

The  fixed  line  OM  is  called  the  directrix  of  the  conic, 
and  the  fixed  point  F,  the  focus.  The  value  of  the  con- 
stant ratio  e  is  called  the  eccentricity.  The  line  perpen- 
dicular to  the  directrix  through  the  focus  is  called  the 
transverse  or  principal  axis  of  the  conic.  The  points  where 
the  transverse  axis  cuts  the  conic  are  the  vertices. 

63.    Parabola.  e  =  l. 

When  e  =  1,  equation  [33]  reduces  to 

7/2  —  2  mx  +  m2  =  0. 

This  curve  has  but  one  vertex,  which  is  evidently  midway 
between  0  and  F.  For  when  y  =  0,  x  =  — .  The  equa- 
tion will  be  reduced  to  a  simpler  form,  if  we  transform  to 
this  point  as  origin.  The  equations  of  transformation  are 
(by  [20]) 

x  =  x  +-9>  and  y  =  y ' 

Substituting  these  values,  the  equation  becomes 

y2  =  2  moc.  [34] 

From  this  equation  we  see  that  the  curve  passes  through 
the  origin;  that  it  is  symmetrical  with  respect  to  the 
X-axis  ;  that  it  is  real  only  to  the  right  of  the  y-axis ; 
and  that  as  x  increases,  y  increases,  —  at  first  more 
rapidly  than  x,  until  x  =  — ,  then  more  and  more  slowly. 
It  has,  therefore,  the  form  shown  in  Fig.   56, 


[02 


ANALYTIC   GEOMETRY 


[Ch.  IX, 


But,  for  the  study  of  the  distant  points,  polar  coordinates 
are  better  adapted.      Transforming  to  polar  coordinates 

with  0  as   origin,   equa- 
tion [34]  becomes 

_  2  m  cos  6 

p~    sin2  e  ' 

"When  6  =  0,  p  is  infinite, 
and  the  curve,  therefore, 
does  not  cut  the  .X-axis 
a  second  time.  But  if 
we  give  to  6  any  finite 
value,  however  small,  p 
will  have  a  finite  value, 
which  will  be  very  large 
for  small  values  of  0,  and  will  decrease  as  6  increases, 
until   for    6  =  — ,  p  —  0.     We    see,   then,   that    every    line 

through  0  except  the  X-ax's  cuts  the  curve  a  second  tune,  a 
fact  which  does  not  appear  from  the  rectangular  form  of 
the  equation.  Yet  the  discussion  of  that  form  showed 
that  the  curve  constantly  recedes  from  the  X-axis.  It  can 
be  shown  by  the  aid  of  the  equation  of  the  tangent  that 
the  curve  approaches  parallelism  with  the  X-axis. 

This  particular  species  of  conic  is  called  the  parabola. 

We  have  already  defined  the  line  iLCVas  the  directrix  ; 
the  point  F  as  the  focus  ;  OX  as  the  principal  axis  ;  and  0 
as  the  vertex  of  the  curve.     We  saw  that 


Fig.   56. 


DO 


\DF 


\m. 


The  coordinates  of  the  focus,  referred  to  0  as  origin,  are 


therefore  f  — ,  0 


771 

The  equation  of  the  directrix  h  x—  —  — , 


Ch.  IX,  §64]  CONIC   SECTION'S  103 

The  line  LI!  through  1\  perpendicular  to  the  -3T-axis 
and  terminated  by  the  curve,  is  called  the  latus  rectum. 

The  abscissa  of  11  is  seen  to  be  — ,  and  by  substituting  this 

value  for  x  in  the  equation  of  the  parabola,  its  ordinate  is 
found  to  be  m.  The  length  of  the  latus  rectum  is  there- 
fore 2  m. 

PROBLEMS 

1.  "What  is  the  equation  of  the  parabola  having  its  vertex 
at  the  origin,  and  its  focus  (a)  on  the  X-axis,  at  a  distance  — 

to  the  left  of  the  origin ;  (b)  on  the  I'-axis,  above  the  origin ; 
(c)  on  the  I"-axis,  below  the  origin  ? 

2.  What  is  the  equation  of  the  parabola  if  the  focus  is  at 

the  origin  and  the  vertex  at  a  distance  —  to  the  left  of  the 
origin  ? 

3.  What  is  the  equation  of  the  parabola,  if  its  vertex  is  at 
the  point  («,  ft)  and  its  axis  is  parallel  to  the  X-axis  ? 

4.  What  is  the  equation  of  the  parabola  which  has  its  ver- 
tex at  the  origin  and  passes  through  the  points  (3,  —  4)  and 
(-3,  -4)? 

5.  Obtain  the  equation  of  the  directrix,  the  coordinates  of 
the  focus,  and  the  length  of  the  latus  rectum  in  the  parabola 

64.    Central  conies.  e^l. 

We  see  from  the  form  of  the  equation  of  a  conic, 

(1  -  e2)x2  -  2  mx  +  y2  +  m2  =  0,  [33] 

that  it  always  represents  a  curve  symmetrical  with  respect 
to  the  .X-axis.  When  e  =  l,  we  have  seen  that  there  is 
but  one  value  of  x  for  each  value  of  y.  But  when  e  =^=  1, 
there  will  be,  in  general,  two  numerically  unequal  values 


104 


ANALYTIC   GEOMETRY 


[Ch.  IX,  §  G4 


of  x  for  any  given  value  of  y.  The  curve  is  therefore 
not  symmetrical  with  respect  to  the  Y"-axis.  But  it  will 
be  shown  to  be  symmetrical  with  respect  to  a  line  parallel 

to  that  axis.  Transform 
the  equation  to  a  new  ori- 
gin midway  between  the 
points  where  the  curve 
— «= — X  cuts  the  X-axis.  The 
Z-axis  will  then  be  found 
to  be  an  axis  of  symmetry. 
Placing  y  =  0  in  [33], 
we  have 


Fig.  57. 


(1  —  e2)x2  —  2  mx  +  m2  =  0. 

The  two  solutions  of  this  equation  will  give  the  inter- 
cepts, OA  and  0Af,  on  the  X-axis.  Let  these  he  denoted 
by  x1  and  x2.  But  we  wish  to  know  00  (=#),  0  being 
the  middle  point  of  A' A. 


Hence 


-      xx  +  x^ 


But  we  know  that  the  sum  of  the  roots  of  a  quadratic 

is  ,  where  a  and  b  are  the   coefficients   of   x2  and  x 

a 

respectively.     (Art.  8.) 


Hence 


xx  +  x2  = 


1-e' 


,  and  x = 


9/1 


The  equations  for  transforming  from  0  to  C  as  origin 
will  then  be  (by  [20]) 


x  —  x'  + 


m 


-,  and  y  =  y' '. 


Ch.  IX,  §64]  CONIC   SECTIONS  105 

Substituting  in  [33],  it  becomes 


^  1         +  l-*2+(l-,2J)2J 


+  y'2  -  2  w*  (  x'  4-  r-^  )  +  m2  =  0 


Reducing  and  dropping  primes, 


Q.-W  +  9 


2  _ 


e1m1 


Dividing  by 


,- 

hn2 

1- 

-e2 
x2 

e2m2 

+-£-=1. 


(1  -  e2)2      1 


e2???2 


Let  — — -  =  r«2,  and  the  equation  becomes 

(1  -  e')2 

2  2 

l+^b]=1-  [35] 

If  then  this  transformation  is  possible,  we  have  reduced 
[33]  to  a  form  which  represents  a  curve  symmetrical  with 
respect  to  both  axes.  It  is  always  possible  except  for 
the  case  when  e  —  L  But  if  e  =  1,  no  value  could  be 
obtained  for  5?,  and  the  point  O  would  not  exist.  This 
has  been  discussed  in  Art.  63.  All  other  cases  are  in- 
cluded in  equation   [35]. 

The  intercepts  of  the  curve  on  the  new  axes,  obtained 
from  equation  [35],  are  ±  a  and  ±  aVl  —  e2.  This  equa- 
tion must  therefore  represent  two  classes  of  curves  quite 
dissimilar  in  form ;  for  while  all  intercepts  are  real  when 
e  <  1,  we  see  that  the  intercepts  on  the  I^-axis  will  be 
imaginary  when  e  >  1.     If,  when  e  <  1,  we  let  the  inter- 


10(3  ANALYTIC   GEOMETRY  [Ch.  IX,  §  65 


cepts  on  the  P"-axis,    ±  a  Vl  —  e2,  be  represented  by  ±  b, 
equation  [35]  becomes 


5+5-1-  ^ 


But  since  we  wish  to  Avork  with  equations  having  only 
real  coefficients,  b  cannot  represent  the  same  expression 
when  e  >  1,  for  Vl  —  e2  would  be  imaginary.  We  then 
let  ±  a  Ve2—  1  =  ±  b' ,  and  equation  [35]  becomes 

%-&-*•  P"] 

The  b  used  in  the  first  case  is  the  actual  intercept. 
In  the  second  case  b'  is  the  real  coefficient  of  the  imagi- 
nary intercept,  and 

b2  =  -  b'\ 

We  see  then  that  there  are  three  distinct  forms  which 
the  locus  may  take.  If  e  =  1,  the  conic  has  been  called  a 
parabola ;  if  e  <  1,  it  is  called  an  ellipse ;  and  if  e  >  1,  an 
hyperbola.  The  ellipse  and  hyperbola  are  called  central 
conies  to  distinguish  them  from  the  non-central  conic,  the 
parabola.  They  may  be  treated  together  from  the  single 
equation  [35],  or  from  their  separate  equations. 

Let  the  student  show  that,  if  the  directrix  is  taken 
as  the  X-axis,  and  a  perpendicular  to  it  through  the 
focus  as  the  J^-axis,  the  simplest  equation  of  the  central 

conies    is    — — —  +  '^-==1.     What  is  its  form  for  the 

«2(1  —  el)      a2, 

ellipse  ?   hyperbola  ? 

65.    Ellipse.  e<l. 

If,  in  Art.  64,  we  had  solved  the  equation 

(l-e2)x2-2mx-\-m2^0, 


Ch.  IX.  §  G5] 


CONIC   SECTIONS 


107 


we  would  have  found  for  the  intercepts  on  the  X-axis, 

x,  =  — ,    and   X%  =  — 

When  e  <  1,  both  these  intercepts  are  positive,  and  there- 
fore both  A  and  A'  lie  to  the  right  of  the  directrix  OY. 
One  intercept,  xv  is  evi- 
dently less  than  m,  and 
therefore  one  point  of 
intersection  is  between 
0  and  F'.     Since  (9(7=     - 


-,  0C>0F\  and  the 


O     ,A  ,F' 


Fig. 


1-e2 

points  must  take  the  po- 
sitions indicated  on  the 
figure. 

We  have  shown  that,  when  e  <  1,  the  equation  of  the 
conic  referred  to  the  new  axes  (see  Fig.  58)  is 


^  +  #1=1 
a2      b*        ' 


[36] 


From  the  form  of  this  equation  we  see  that  the  curve  is 
symmetrical  with  respect  to  both  axes,  and  hence  to  their 
intersection ;  that  it  cuts  the  X-axis  at  the  points 
(±  a,  0),  and  the  F-axis  at  (0,  ±  b)  ;  that  the  values  of  x 
are  real  only  for  values  of  y  from  —  b  to  +  b ;  and  that 
the  values  of  y  are  real  only  for  values  of  x  from  —  a  to 
-f  a.  A  more  careful  plotting  of  the  points  will  show 
that  it  has  the  form  shown  in  Fig.   59. 

The  line  D'lP  has  been  called  the  directrix,  and  the 
point  Ff  the  focus.  Place  the  points  F  and  D  on  the 
X-axis  so  that  CF=F'Q  and  CD  =  D'C,  and  draw  DR 
perpendicular  to  the  X-axis.     The  symmetry  of  the  curve 


108 


ANALYTIC    GEOMETRY 


[Ch.  IX,  §  65 


shows  that  if  we  had  used  the  line  BH  and  the  point  F  as 
directrix  and  focus,  and  the  same  value  of  e,  the  same 
curve  would  have  been  found  as  the  locus.  The  curve 
can  be  said  therefore  to  have  these  two  lines  BR  and 
B'R'  as  directrices,  and  the  two  points  F  and  F'  as  foci. 


We  can  now  obtain  the  relations  between  the  various 
lines  in  the  figure.     We  have  seen  that 

FB  =  D'F'  =  m, 
CB  =  B'C  = 


1-e5 


CA  =  A'C  =a 


em 


CB  =  B'0  =b  = 


em 


VI 


It  follows  that  CB  =  ~,  and  that  the  equations  of  the 
directrices  are 

v  =  %  and  x=-^.  [38] 

e  e 


Ch.  IX,  §  66)  CONIC   SK(TI()XS  1U(.) 

Also  that        CF=CD-FI)  =  — ^—  -  wi  =  -^-  =  ae. 

1   —  £2  l  —  ^2 

It  is  convenient  to  let  CF  be  represented  by  a  single 
letter  c. 


Then 


ae,  or    ^ 


In    obtaining    equation    [36],    we  let   b2  =  a2(l  —  e2~). 

Solving  for  e2,  we  have 

*  =  «L=JL  [39] 

a2  u     J 


Comparing  these  two  values  of  e,  we  have 
a2  —  b2  =  c2. 


[40] 


From  this  we  see  that  BF,  being  the  hypotenuse  of  a 
right  triangle  whose  legs  are  c  and  5,  is  equal  to  a.     It 
also     shows     that    a     is 
always  larger  than  b,  or 
that      i'i(=2a),      the 
axis  perpendicular  to  the 
directrices,  is  larger  than 
B'B(=2b).      A' A     has 
been     called    the    trans-    x~ 
verse   or   principal   axis ; 
B'B  is  called  the  conju- 
gate or  minor  axis  of  the 
curve. 

If  the  foci  of  the  ellipse 
are  on  the  I^-axis,  the 
vertex  A  also  lies  on  that 

axis,  and  B  on  the  .X-axis  (Fig.  60).     Its  equation  is  (see 
end  of  Art.  64) 

-  +  ^=1.  [41] 

b2      a2  L     J 


110  ANALYTIC   GEOMETRY  [Ch.  IX,  §  GO 

All  the  formulas  found  above  hold  for  [41],  except  the 
equations  of  the  directrices,  which  are 

,  a 

y  =  ±- 

e 

PROBLEMS 

1.  Find  a,  b,  c,  e,  and  the  equations  of  the  directrices  in  the 
ellipse, 

(a)  4a2  +  9?/2  =  36,  (b)  9x*  +  4:y2  =  ?>6, 

(c)  3ar°  +  8>/2=10. 

2.  Find  the  equation  of  the  ellipse  having  its  centre  at  the 
origin  and  its  foci  on  the  AT-axis,  if 

(a)  a  =  3  and  b  =  2,  (d)  b  =  4  and  c  =  3, 

(&)    b  =  3  and  e  =  i,  (e)   a  =  5  and  c  =  3, 

(c)    a  =  6  and  e  =  f,  (/)  c  =  4  and  e  =  J. 

3.  Show  that  the  length  of  the  latus  rectum  (line  through 

2  b- 
the  focus  perpendicular  to  the  axis)  of  the  ellipse  is 

4.  Show  that  the  circle  is  the  limiting  form  of  the  ellipse 
as  a  and  b  approach  equality.  What  is  the  eccentricity  of  the 
circle,  and  where  are  its  foci  and  directrices  ? 

5.  What  is  the  equation  of  the  ellipse  which  has  its  centre 
at  the  origin  and  its  axes  coincident  with  the  coordinate  axes, 
and  which  passes  through  the  points  (4, 1)  and  ( —  3,  2)  ? 

6.  What  is  the  equation  of  an  ellipse  if  its  centre  is  at  the 
point  (a,  /3)  and  its  axes  are  parallel  to  the  coordinate  axes  ? 

66.    Hyperbola.  e>l. 

When   e  >  1,   one  of  the   intercepts, -,  is  positive, 

1  +  e 

071 

and  less  than  ??i,  while  the  other, ,  is  negative.      OC 

1  —  e  ° 

will  also  be  negative.     The  points,  A,  A',  (7,  and  .F,  will, 
therefore,  take  the  positions  indicated  in  Fig.  61. 


Cii.  IX,  §  66] 


('••NIC    SECTIONS 


111 


We  have  shown  that,  when  e  >  1,  the  equation  of  the 
conic  reduces  to 


A' 


AF 


--^-  =  1.    r37i 

a2      b'2  L     J 

Again  we  see  that  the    - 
curve  is  symmetrical  with 
respect  to  hoth  axes,  and 
hence  with  respect  to  the 

origin;  that  it  cuts  the  X-axis  at  the  points  (±  a,  0),  and 
does  not  cut  the  !F-axis ;  and  that  the  values  of  y  are  real 
only  for  values  of  x  numerically  equal  to  or  greater  than  a. 
The  exact  form  can  be  obtained  more  readily  from  the  polar 


Fig.  61. 


equation.     Transforming  — 

a1 

with  C  as  origin,  we  have 


22—  =  1  to  polar  coordinates 

b'2  l 


P2  = 


a2b'2 


&'2COs20-a2sin20 

When  6  =  0,  p  =  ±  a,  and  as  6  increases,  the  denomi- 
nator decreases,  the  fraction  increases,  and  the  point 
recedes  from  the  origin.  This  will  continue  as  long  as 
the  denominator  remains  positive.  As  soon  as  the  de- 
nominator becomes  negative,  the  value  of  p  becomes 
imaginary.  There  is  then  a  value  of  0  beyond  which 
the  curve  does  not  exist.  This  value  of  6  is  that  which 
makes  the  denominator,   b'2  cos2  6  —  a2  sin2  0,  zero,  or 

a 
For    every    value     of     6    between 
tan 


6  =  tan" 


tan"1! 


and 


there  will  be   a  real  value   of  p,  these  val- 
ues    growing    larger    as    6    approaches    tan_1(H — j 


112 


ANALYTIC   GEOMETRY 


[Ch.  IX,  §  66 


tan-1( ).     The    lines   then    which   pass    through    the 

V     aJ                                     f     V\                     f     V\ 
origin,  making   the  angles  tan  M  H J  and     tan_1( J 

with  the  JT-axis,  do  not  cut  the  curve,  while  every  line 
lying  between  these  lines  cuts  the  curve  in  two  real  points. 
The  curve  must  therefore  approach  parallelism  with  these 
lines  as  the  point  recedes  from  the  origin,  and  it  will  be 
shown  in  the  next  article  that  the  curve  approaches  coinci- 
dence with  these  lines.  Such  a  line  is  called  an  asymptote. 
If  we  continue  to  increase  0,  we  see  that  there  will  be 

no  real  value  oi.p  until  tan  6  again  becomes  numerically 

b> 
less  than  —      Then   p   goes    through   the    same   changes 
a 

in  value,  decreasing  until  it  equals  ±  a.  But  we  have 
shown  that  the  curve  is  symmetrical  with  respect  to  both 
axes,  and  there  is  therefore  no  need  of  discussion  beyond  the 
first  quadrant.     The  following  is  the  form  of  the  hyperbola : 

7 


Place    the    points   F'   and  D'   on    the    X-axis   so    that 
F'C=OF  md  D'C=CD,  and  draw  D'H'  perpendicular 


Ch.  IX,  §  GG]  CONIC   SECTIONS  113 

to  the  X-axis.  The  symmetry  of  the  curve  again  shows, 
as  in  the  ellipse,  that  the  hyperbola  may  be  said  to  have 
two  foci,  J7  and  F\  and  two  directrices,  DR  and  D' W '. 

We  can  now  <>l>!lfc^tln'  relations  between   the  various 
lines  in  the  figure.   ^Bliavc  seen  that 

DF=FD'  =  m, 


e*-r 


CA  =  A'C  =a  = 


em 


e2-l 


CB  =  B'0  =b'  =— — 


VeP-1 


It  follows  that   CD  =  -,  and  that  the  equations  of  the 
directrices  are 


a  t  a 

ac  = 

Also  that 


*  =  2,  and   *  =  -?.  ,    [42] 


0F=  CD  +  DF=^-+m  =  -pL  =  ae. 
ez  —  1  e&  —  1 

It  is  convenient  to  let  OF  be  represented  by  a  single 
letter  c. 

Then  c  =  ae, 

e 

or  e  =    . 

a 

In    obtaining    equation    [37],    we    let    V2  =  a2(e2  —  1). 

Solving  for  e2,  we  have 

Comparing  the  two  values  of  e,  we  have 

a?  +  b'2  =  c2.  [44] 


114  ANALYTIC   GEOMETRY  [Ch.  IX,  §  G7 

There  is,  in  the  hyperbola,  no  restriction  on  the  relative 
values  of  a  and  b' . 

Note.  —  In  the  following  articles  we  shall  follow  the  ordinary  custom, 
and  use  b  in  place  of  b'. 


s    f  i    til 


67.  Asymptotes.  —  The  slopes  of  the  asymptotes  were 
seen  [Art.  60]  to  be  ± -.     Hence  their  equations  are 

b  .  b 

y=~x->  and  y  =  ~ax> 

or  written  as  a  single  equation, 

a2      b2 

They  are  evidently  the  diagonals  of  the  rectangle  formed 
by  drawing  lines  parallel  to  the  axes  through  A,  A',  B, 
and  B'. 

It  remains  to  be  shown  that  the  curve  not  only  approaches 
parallelism  with  these  lines,  but  actually  approaches  coinci- 
dence with  them  ;  or  that  the  perpendicular  distance  PXM 
from  any  point  Px  on  the  hyperbola  to  the  asymptote 
decreases  indefinitely,  as  Px  recedes  from  the  origin  along 
the  curve.     (See  Fig.  62.) 

Since   the   equation    of   the    asjmiptote   is   bx  —  ay  =  0, 

P1M=  bx*  -  ay±-  (By  [17]) 

V62  +  a2 

But,  since  Px  is  a  point  on  the  curve, 

b2xx2  —  a2yx2  =  a2b2, 
or,  factoring, 

bxx  -  ayx 


bxx  +  ayx 


Ch.  IX,  §  G7]  CONIC   SECTIONS  115 

Hence  1\M 


(^1  +  a?/1)V^  +  a2 

This  expression  evidently  decreases  as  x1  and  yx  in- 
crease, approaching  zero  as  a  limit.  The  curve  therefore 
a pp roaches  its  asymptote  indefinitely, 

PROBLEMS 

1.  Find  a,  b,  c,  e,  and  the  equations  of  the  directrices  and 
asymptotes  of  the  hyperbola, 

(a)  a*-25tf  =  25,      0)  9x2--if-  =  3(j,       (c)  2x2-5f-  =  20. 

2.  Find  the  equation  of  the  hyperbola  having  its  centre  at 
the  origin  and  its  foci  on  the  X-axis,  if 

(a)  a  =  3  and  b  =  2,  (d)   b  =    4  and  c  =  5, 

(b)  b  =  3  and  e  =  2,  (e)    a  =   4  and  c  =  5, 

(c)  a  —  5  aild  e  =  f ,  (/)  c  =  10  and  e  =  3. 

3.  What  is  the  equation  of  an  hyperbola,  if  its  centre  is  at 
the  point  («,  /?)  and  its  axes  are  parallel  to  the  coordinate  axes  ? 

4.  What  is  the  equation  of  the  hyperbola  which  has  its 
centre  at  the  origin  and  its  foci  on  the  X-axis,  and  which 
passes  through  the  points  (5,  3)  and  (—3,  2)  ? 

5.  Show  that  the  latus  rectum  of  the  hyperbola  is  - — • 

6.  Show  that  the  foot  of  the  perpendicular  from  the  focus 
of  an  hyperbola  on  an  asymptote  is  at  the  distance  a  from  the 
centre  and  b  from  the  focus. 

7.  Show  that  the  circle  of  radius  b,  whose  centre  is  at  the 
focus  of  an  hyperbola,  is  tangent  to  the  asymptote  at  the  point 
where  it  is  cut  by  the  directrix. 

8.  Show  that  the  product  of  the  two  perpendiculars  let  fall 
from  any  point  of  an  hyperbola  on  the  asymptotes  is  constant. 


116 


ANALYTIC   GEOMETRY 


[Ch.  IX,  §  68 


68.  Conjugate  hyperbolas.  —  Tf,  in  deriving  the  equa- 
tion of  the  conic,  the  directrix  is  taken  as  the  X-axis,  and 
a  perpendicular  to  it  through  the  focus  as  the  JT-axis,  its 


Fig.  03. 
simplest  form,  in  the  case  of  the  hyperbola,  is 

=  1. 


#2      y2  __ 


If  the  definitions  of  a  and  b  are  interchanged,  using  b  to 
represent  the  semi-transverse  axis  (which  is  here  the  real 
intercept  of  the  hyperbola  on  the  !F-axis),  the  equation 
becomes 


</2_ 


/;- 


1. 


The  hyperbolas  —  —  *-  =  1   and  —  —  %—  =  —  1,  where  a 
a2      ¥  a2      bl 

and   b  have  the   same   values,   are   closely   related.     The 
transverse  and  conjugate  axes  of  the  first  are  respectively 


Ch.  IX.  §69]  CONIC   SECTIONS  111 

the  conjugate  and  transverse  axes  of  the  second.  Two 
hyperbolas  which  are  so  related  arc  called  conjugate 
hyperbolas,  either  being  conjugate  to  the  other.  Bui  it 
is  convenient  to  speak  of  the  first  as  the  primary  and  the 

second  as  the  conjugate  hyperbola. 

The  polar  equation  of  the  conjugate  hyperbola, 

-  a%2 


P2  = 


52  cog2  0  _  a2  sin2  0 


differs  from  that  of  the  primary  hyperbola  only  in  the  sign 
of  the  second  member.  It  therefore  gives  real  values  for 
p  only  for  those  values  of  6  which  gave  imaginary  values 
for  p  in  the  primary  hyperbolas.  The  conjugate  hyperbola 
has  therefore  the  same  asymptotes  as  the  primary,  but 
is  situated  on  the  opposite  sides  of  those  asymptotes. 

The  value  of  c  ( =  Va2  +  62)  is  the  same  for  both  the 
primary  and  conjugate  hyperbolas,  and  the  four  foci  there- 
fore lie  on  a  circle  having  its  centre  at  the  origin.  But 
for  the  conjugate  hyperbola  er  =  7,  and  the  equations  of 
the  directrices  are  y  —  ±  —  • 


e 


69.    Equilateral    or    rectangular    hyperbola. — If   b  =  a, 

the  equation  of  the  hyperbola  becomes  x2  —  y2  =  a2.  This 
is  called  the  equilateral  hyperbola.  The  equations  of  its 
asymptotes  are  x  +  y—  0  and  x  —  y  =  0.  They  therefore 
make  an  angle  of  45°  with  the  X-axis,  and  are  perpen- 
dicular to  each  other.  From  this  fact  it  is  often  spoken 
of  as  the  rectangular  hyperbola. 

The  equilateral  hyperbola  and  its  conjugate  are  evi- 
dently equal  to  each  other.  Figure  63  shows  an  equi- 
lateral hyperbola  and  its  conjugate. 


118 


ANALYTIC   GEOMETRY 


[Ch.  IX,  §  70 


PROBLEMS 

1.  Write  the  equation  of  an  hyperbola  conjugate  to  the 
hyperbola  4  a,-2  —  y2  =  4,  and  find  its  axes,  eccentricity,  latus 
rectum,  the  coordinates  of  its  foci,  and  the  equations  of  its 
directrices. 

2.  Find  the  eccentricity  of  the  equilateral  hyperbola. 

3.  Show  that  if  e  and  e'  are  the  eccentricities  of  two  conju- 
gate hyperbolas,  —  +  —  =  1  and  ae  =  be'. 

4.  Find  the  equation  of  the  equilateral  hyperbola  referred 
to  its  asymptotes  as  axes. 

Note.  — Revolve  the  axes  through  the  angle  —  45°. 


70.  Focal  radii  of  a  central  conic.  —  The  distance  of  a 
point  on  a  conic  from  a  focus  is  called  a  focal  radius  of  the 
point.  Since  there  are  two  foci  in  a  central  conic,  there 
will  be  two  focal  radii  for  each  point  of  the  conic. 

The  distance  FPX  of 
any  point  (xv  y^)  of  the 
ellipse  from  the  right- 
hand     focus    (ae,  0)     is 

(by  [l]) 


V  (x-x  —  ae)2  +  #!2. 

But  since  (xv  y±)  is  a 
point  on  the  ellipse 


Fig.  64. 


b2x2  +  a2y2  =  a2b2, 
its  coordinates  must  satisfy  this  equation. 

Hence 


Vx^^-T^ 


a? 


Oh.  IX,  §  70] 


CONIC   SECTIONS 


119 


Substituting  this  value  of  y^  in  the  expression  for  the 
distance,  Ave  have 


FP1  =  yjx2  -  2  aexx  +  a2e2  +  b2  -  h\  x2. 

a2  —  b2 

Noting  that —  =  e2,  and  a2e2  +  b2  =  a2, 

a2 

this  reduces  to 


FPX  =  Va2  -  '2  aex1  +  e2x2  =  ±(a-  exj. 

The  distance,  F'PV  of  the  point  from  the  left-hand 
focus  may  be  found  in  the  same  way  except  that  the  coor- 
dinates of  F'  are  (—  ae,  0). 

Hence  F'Pl  =  ±  (a  +  ex{). 

Let  the  student  show  that,  though  the  work  in  the  case 
of  the  hyperbola  will  be  slightly  different,  the  results  will 
be  the  same. 

Since  it  is  only  the  length  of  the  focal  radii  that  we 
seek,  it  will  be  neces- 
sary to  determine  in 
each  conic  which  sign 
should  be  used  before 
the  parenthesis,  so  that 
it  may  express  a  posi- 
tive distance.  In  the 
ellipse     a     is     always 

greater  than  exv  and  the  positive  sign  must  therefore  be 
used  in  both  cases. 

Hence,  in  the  ellipse, 

FPX  =  a-  exu 
and  F'P1  =  a  +  exu 


[45] 


120  ANALYTIC   GEOMETRY  [Ch.  IX,  §  71 

It  will  be  necessary  to  consider  the  two  branches  of  the 
hyperbola  separately.  For  the  right-hand  branch  exx  is 
always  positive  and  greater  than  a;  and  the  two  dis- 
tances are 

FP-.  =  exi  -  a, 

[46,  »] 
and  F'P1  =  exi  +  a. 

For  the  left-hand  branch  ex1  is  negative  and  greater  in 
absolute  value  than  a ;   and  the  two  distances  are 

FP,  =  —  exi  +  a, 

[46,  .5] 
and  FrP1  =  —  exx  -  a. 

From  these  results  we  see  that  the  sum  of  the  two  focal 
radii  of  any  point  of  an  ellipse  is  2  a.  While  in  the  hyper- 
bola the  difference  of  the  tiro  focal  radii  is  2  a.  The 
ellipse  might  therefore  be  defined  as  the  locus  of  points, 
the  sum  of  whose  distances  from  two  fixed  points  is  con- 
stant ;  and  the  hyperbola  as  the  locus  of  points,  the 
difference  of  whose  distances  from  two  fixed  points  is 
constant. 

Let  the  student  obtain  the  equations  of  the  ellipse  and 
hyperbola  in  their  ordinary  forms  from  these  definitions. 

71.  Mechanical  construction  of  the  conies.  — The  results 
of  the  last  section  enable  us  to  construct  mechanically  the 
ellipse  and  hyperbola. 

To  construct  the  ellipse  fix  two  pins  at  the  foci,  and 
place  around  them  a  loop  of  string  whose  length  is 
2a  +  2c.  If  a  pencil  is  placed  in  the  loop  and  moved 
about  the  foci,  keeping  the  string  taut,  it  will  describe  an 
ellipse.     The  proof  is  evident. 


Ch.  IX,  §  72] 


CONIC   SECTIONS 


121 


Fig.  66. 


An  hyperbola  may  be  traced  in  the  following  manner 
Fix  one  end  of  a  ruler  at 
one  focus,  F' .  A  string 
whose  length  is  2  a  less  than 
the  length  of  the  ruler  is 
attached  to  the  focus  F  and 
to  the  other  end  of  the  ruler. 
A  pencil,  which  holds  the 
string  taut  and  against  the  ruler,  will  trace  a  branch  of 
the  hyperbola.     For  in  all  positions  of  P, 

F'P  -FP=2a. 

The  parabola  is  per- 
haps more  easily  traced 
by  points.  Erect  a  per- 
pendicular ML  at  any 
point  of  the  axis.  With 
F  as  a  centre  and  a  radius 
equal  to  DM,  describe  an 
arc,  cutting  ML  at  P. 
Then  P  is  a  point  of  the 
parabola.  For  JSTP  =  FP. 
As    many  points    as  we 

please  may  be  found  in  this  way  and  the  parabola  passed 

through  them. 

72.  Auxiliary  circles.  —  The  auxiliary  circle  of  a  cen- 
tral conic  is  a  circle  described  on  the  major  axis  as 
diameter.     Its  equation  is  x2  +  y2  =  a2. 

The  circle  described  on  the  minor  axis  as  diameter 
is  called  the  minor  auxiliary  circle.  Its  equation  is 
x2  -f  y2  =  b2. 


Fig.  67. 


122 


ANALYTIC   GEOMETRY 


[Ch.  IX,  §  72 


Points  on  the  ellipse  and  auxiliary  circle  which  have 
the   same    abscissa   are    called   corresponding    points.      In 


Fig.  68. 

Fig.  68,  Px  and  R  are  corresponding  points.     The  angle 
MXCR  is  caHed  the  eccentric  angle  of  the  point  Pv 

There  is  a  simple  relation  between  the  ordinates  of  the 
corresponding  points  P1  and  R  which  may  be  found  in 
the  following  manner.  Let  the  coordinates  of  Pl  be 
(xv  y^,  and  of  R  (xv  ?/2). 

Substituting  these  values  for  x  and  y  in  the  equations 
of  the  ellipse  and  circle  respectively,  we  have 

b2x^  +  a2y12  =  a2b2, 
and  x?  -f-  y2  =  a2. 

Multiplying  the  second  equation  by  b2  and  subtracting, 
we  have 


or 


a2y2  =  b2y2, 

v± = ±  _. 

y2        a 


Or,  the  ordinate  of  any  point  of  the  ellipse  is  to  the  ordinate 
of  the  corresponding  point  of  the  circle  as  b  is  to  a. 


Ch.  IX,  §  73]  CONIC   SECTIONS  123 

Let  the  student  show  that  a  similar  relation  holds 
between  the  abscissas  of  points  on  the  ellipse  and  minor 
auxiliary  circle  which  have  the  same  ordinate.  These  are 
also  called  corresponding  points.  Show  that  CR  passes 
through  the  corresponding  point  in  the  minor  auxiliary 
circle. 

PROBLEMS 

1.  Find  the  focal  radii  of  the  ellipse  ar  +  9  y2  =  18  for  the 
points  whose  abscissa  is  —  2. 

2.  Find  the  focal  radii  of  the  hyperbola  9  v?  —  4  y2  =  65  for 
the  points  whose  ordinate  is  2. 

3.  Show  that  the  distance  of  a  point  on  an  equilateral 
hyperbola  from  the  centre  is  a  mean  proportional  between  the 
focal  radii  of  the  point. 

4.  Prove  that  the  circle  described  on  any  focal  radius  of  an 
ellipse  as  a  diameter  is  tangent  to  the  auxiliary  circle. 

5.  Prove  that  the  auxiliary  circle  of  an  hyperbola  passes 
through  the  intersections  of  the  directrices  and  asymptotes. 

6.  Show  that  the  focal  radius  of  any  point  of  a  parabola  is 

1      2 

7.  Show  that  the  area  of  the  ellipse  is  -n-ab. 

Note.  — Divide  the  major  axis  of  the  ellipse  into  any  number  of  equal 
parts,  and  on  each  of  these  parts  inscribe  rectangles  in  the  ellipse  and 
auxiliary  circle.  The  areas  of  these  rectangles  will  be  in  the  ratio  b  :  a, 
and  by  the  theory  of  limits  it  may  be  shown  that  the  areas  of  the  ellipse 
and  auxiliary  circle  will  be  in  the  same  ratio. 

73.  General  equation  of  conies  when  axes  are  parallel  to 
the  coordinate  axes.  —  From  the  equations  of  the  conies 
which  have  been  determined,  we  may  obtain  by  transfor- 
mation of  coordinates  the  most  general  form  of  their 
equations  referred  to  any  set  of  axes.     A  full  discussion 


124  ANALYTIC    GEOMETRY  [Ch.  IX,  §  73 

of  this  question  will  be  taken  up  in  Chapter  XIII.,  where 
it  will  be  shown  that  every  equation  of  the  second  degree 
represents  some  form  of  the  conic.  For  the  present  we 
shall  content  ourselves  with  a  very  short  discussion  of 
their  equations  when  the  coordinate  axes  are  parallel  to 
the  axes  of  the  conic.  When  this  is  the  case  and  the 
coordinates  of  the  centre  are  («,  /3),  the  equations  of  the 
central  conies  take  the  form. (by  [20] ) 

(x-a)2      Q/-/3)2 
a2  b* 

The  parabola  whose  vertex  is  at  the  point  («,  y8),  and 
whose  axis  is  parallel  to  the  Jf-axis,  takes  the  form 

(y-/3)2  =  ±2™<>-«). 

If  the  axis  of  the  parabola  is  parallel  to  the  Y"-axis,  its 

equation  is 
*  0-«)2=±2m(j/-/3). 

In  each  case  the  term  in  xy  is  wanting,  and  all  of  the 
equations  are  seen  to  be  special  cases  of  the  general 
equa  ion  ^  +  ^  +  J)x  +  ^  +  F  =  ^ 

If  neither  A  nor  C  is  zero  in  this  equation,  it  may  be 
written  in  the  form 

aU  +  v-rr-^icfv*  +  Ev+:E2  V  m  +  &  - f- 

A\ X+A  X  +  ±A*)  +  tV  +Cy  +  4Cr>riA  +  IC  F' 
or,  if  we  represent  the  second  member  by  K, 


(•+n)   (»  + 


2  0 


+  v        "       =1. 


K  K 

A  O 


Ch.  IX,  §73]  CONIC  SECTIONS  125 

If  A  and  C  have  the  same  sign,  this  takes  the  form  of 
the  equation  of  the  ellipse  whose  centre  is  at  the  point 

and  in  which  a  —  \ — ,  and  b=\—-.     The 


-1A     26V  _^  *0 

ellipse  wiH  be  real,  null,  or  imaginary,  according  as  a  and 

b  are  real,  zero,  or  imaginary. 

Let  the  student  show  that  if  A  and  0  have  opposite 
signs,  the  equation  represents  an  hyperbola,  or  (if  K  =  0) 
two  intersecting  lines. 

Also  that,  if  either  A  or  0  is  zero,  the  equation  repre- 
sents a  parabola,  or  a  pair  of  parallel  lines. 

PROBLEMS 

1.  Determine  the  nature  and  position  of  the  locus  of 

2x2  +  3y2-6x  +  4:y  =  l0. 
This  equation  may  be  written  in  the  form 

2(^-3x  +  f)  +  3(.!/+|2/  +  |)  =  10  +  f  +  |=^) 

(*-f)2 ,  (y  +  f)2,-. 

9  5  T  9  5 

12  18 

The  locus  is  an  ellipse,  having  its  centre  at  the  point 
(f ,   —  -§),  and  in  which  a  =  Vff ,  and  b  =  V-f-f • 

2.  Determine  the  nature  and  position  of  the  locus  of  the 
following  equations : 

(a)  x>  +  2y2-6x  +  y  =  10,  (d)  3ar  -  y2  +  Gy  =  0, 

0)  a?  +  4x-2y  =  W,  (e)   f  +  2x-±y  =  6 

(c)  ±x2-3if-±x  +  S  =  0, 

3.  Obtain  the  polar  equation  of  each  of  the  conies,  the  focus 
being  used  as  the  origin  and  the  transverse  axis  as  the  initial 
line. 


CHAPTER   X 


TANGENTS 

74.  The  method  of  finding  the  equation  of  a  tangent  to 
any  conic  at  a  given  point  is  the  same  as  that  used  in  the 
case  of  the  circle  (Art.  55}.  The  equation  of  a  secant 
through  the   given   point  Pv   (xv  y±),  and  an  adjacent 

point  P2,  (xx  +  h,  yx  +  &), 


on  the  curve  is 

V  ~  V\  _  * 
x  —  x1      h 

It  is  necessary 
termine  in  each 

to  de- 
case  a 

Fig.  69. 


value   of   -  which  will 
h 

not     be     indeterminate 
when  h  and  k  approach 

zero.     We  shall  give  the  work  in  detail  for  the  ellipse, 

b2x2  -\-  a2y2  =  a2b2. 

Since   the  points  Px   and  P2  lie   on  the   ellipse,  their 

coordinates  must  satisfy  its  equation,  or 

(1)  b2x2  +  a2y2=a2b\ 

(2)  b2x2  +  2  bVix^  +  b2h2  +  a2y2  +  2  a2kyx  +  a2k*  =  a2b2. 
Subtracting  (1)  from  (2),  we  have 

2  b2hxx  +  b2h2  +  2  a2kyx  +  a2k2  =  0, 

Jc=      2  b2x,  +  b2h 
h  2  a2yx  +  a2k 

126 


or 


Ch.  X,  §  74]  TANGENTS  127 

The  equation  of  the  secant  may  therefore  be  written 

//  -  ,yt  =       2  b2xx  +  b2h 
x  —  x1  2  cfiyl  +  a2k 

Now,  if  we  let  P2  approach  Pv  h  and  &  will  approach 

zero,  and   the   limit   of  the  second  member  is  no  longer 

b2x 
indeterminate,  but  becomes 1- 

The  equation  of  the  tangent  is  therefore 

y-y\=    ft2*i7 

x  -  xx  a2yx 

or  clearing  of  fractions  and  transposing, 

b2xxx  +  a2y^y  =  b2x2  +  a2y2. 

But  b2x2  +  a2y2  =  a2b\ 

and  the  equation  of  the  tangent  reduces  to 

b-xix  +  aryxy  =  a2b2.  [47] 

Let  the  student  show  that  the  equation  of  the  tangent  to 
the  hyperbola, 

b2x2  -  ahf  =  a2b2,    is    bxxx  -  cpyxy  =  aW-,         [48] 

the  parabola,        y2=2mx,  is    y\y  =  m(x  +  xi),  [^9] 

the  locus  of  the  general  equation  of  the  second  degree, 

Ax2  +  Bxy  +  Cy2  +  Dx  +  Ey  +  F=  0, 

is  Axix+~-(x{y  +  yix)+  Cyxy 

[50] 

+  ~(x  +  xl)  +  §(y  +  yO+F=o. 

These  formulas  can   be  most   easily  remembered   and 
applied  if  we  notice  that  they  may  be  obtained  from  the 


128  ANALYTIC   GEOMETRY  [Ch.  X,  §  75 

equation  of  the  conic  by  replacing  x2  and  y2  by  xxx  and 

constant  quantities  being  unchanged. 

The  method  of  finding  the  equations  of  the  tangents 
from  an  exterior  point  is  the  same  as  that  given  for  the 
circle.     (See  Art.  57.) 

Let  the  student  show  that  the  equation  of  the  chord  of 
contact  of  tangents  from  an  exterior  point  will,  in  each 
case,  take  the  same  form  as  the  equation  of  a  tangent  at 
the  point  of  contact.     (See  Art.  59.) 

75.  Normals.  —  The  normal  at  any  point  of  a  conic  is 
the  line  through  the  point,  perpendicular  to  the  tangent 
at  the  point. 

It  can  be  found  in  any  case  by  writing  the  equation  of 
a  tangent,  and  then  writing  the  equation  of  a  perpendicu- 
lar to  the  tangent  through  the  point  of  contact. 

For  example,  the  tangent  to  the  ellipse  has  been  found 
to  be  b2xxx  +  a2yxy  =  a2b2.  A  perpendicular  to  this  line 
will  have  the  form  a2yxx  —  b2xYy  =  k.  Since  the  normal 
passes  through  Pv  k  =  cfiy^^  —  b2x^jv  and  the  equation 
of  the  normal  becomes 

a2yxx  -  b2xxy  =  (a2  -  b2)x1yv 

In  like  manner,  the  equation  of  the  normal  to  the 
hyperbola  is      ^  +  ^  =  ^  +  ^ 

and  to  the  parabola  is 

yxx  +  my  =  x1yl  +  myv 

The  student  should  note  that  these  formulas  apply  only 
when  the  equations  of  the  curves  are  in  the  simplest  form. 


Cn.  X,  §  7G] 


TANGENTS 


129 


He  is  advised  not  to  use  them  in  solving  problems,  as 
they  are  not  easily  remembered,  but  in  each  case  to  write 
the  equation  of  the  tangent  and  then  that  of  a  perpen- 
dicular to  the  tangent  at  the  point  of  contact. 

76.  Subtangents  and  subnormals. — The  projections  on 
the  X-axis  of  those  parts  of  the  tangent  and  normal 
included  between  the  point  of  contact  and  the  X-axis 
are  called  the  subtangent  and  subnormal. 


Eig.  70. 

In  the  ellipse  (Fig.  70)  MXT  is  the  subtangent  and 
JfjiV  is  the  subnormal  for  the  point  Pv  The  equation 
of  the  tangent  at  P1  is  b2xxx  -f  a2yxy  =  a2b2,  and  the  equa- 
tion of  the  normal  is  cPy1x—b2xxy  —  (cP'  —  b2:)x1yl.  Find- 
ing the  intercepts  on  the  X-axis,  we  have 


CT=~    and   CN=  ^  ~  ^  xv 


But 


MXT=  CT-  CMX  =  - 


ocf 


OCi 


and    M^^Ctf-CM^ 


(a2  -  b2)x^ 


V 


Xv 


[51] 

[52] 


130  ANALYTIC  GEOMETRY  [Ch.  X,  §  70 

2  2 

Let  the  student  show  that  for  the  hyperbola  —  —  &-  =  1, 

a2      b2 

the  subtangent  equals  a  ~  a?1  >  [53] 

the  subnormal  equals  -^«i,  [54] 
and  that  for  the  parabola  y2  =  2  mx, 

the  subtangent  equals   -  2  xi,  L^li 

the  subnormal  equals  in,  [5t>] 

PROBLEMS 

1.  Find  the  equations  of  the  tangents  and  normals  to 

(a)  3ar9  +  4?/2  =  l9  at   (1,  2), 

(6)  2ar*-2/2  =  14  at  (3,-2), 

(c)    ?r  =  6  ;i'  at  (6,  —  6), 

(t?)  2^2-3a7/  +  6.T-2  =  0  at  (2,3). 

2.  Find  the  lengths  of  the  snbtangents  and  subnormals 
in  (a),  (6),  and  (c),  problem  1. 

3.  Find  the  equations  of  the  tangents  to 

(a)  16  x2  +  25  y2  =  400  from  (3,  4), 

(b)  y-  =  ±x  from  (-3,  -2), 

(c)  ar- 3/ +2  a- +  19  =  0  from  (-1,2). 

4.  Find  the  chords  of  contact  of  the  tangents  in  problem  3. 

5.  Find  the  lengths  of  the  tangents  and  normals  in  («),  (Jj), 
and  (c),  problem  1. 

Note.  —  The  terms  "length  of  tangent"  and  "  length  of  normal1'  are 
used  to  indicate  the  distances  on  the  tangent  and  normal  from  the  point 
of  contact  to  the  points  where  they  cut  the  X-axis. 

6.  Find  the  angles  between  the  ellipse  4  x2  +  y2  =  5,  and 
the  parabola  y2  =  8  x,  at  their  points  of  intersection. 

7.  Show  how  the  subtangent  or  subnormal  in  the  parabola 
may  be  used  to  construct  the  tangent  at  any  point  of  the  curve. 


Ch.  X,  §  77]  TANGENTS  131 

8.  From  the  fact  that  the  subnormal  in  the  parabola  is 
constant,  show  that  the  tangent  approaches  parallelism  with 
the  axis  as  the  point  of  contact  recedes  from  the  origin. 

9.  Show  that  if  the  normals  of  an  ellipse  pass  through 
the  centre,  the  ellipse  is  a  circle. 

10.  Show  that  the  distance  from  the  focus  of  a  parabola 
to  any  tangent  is  one-half  the  length  of  the  corresponding 
normal. 

11.  Show  that  the  focus  of  a  parabola  bisects  the  portion 
of  the  axis  intercepted  by  a  tangent  and  the  corresponding 
normal. 

77.  Slope  form  of  the  equations  of  tangents.  —  For  many 
problems  it  is  convenient  to  have  the  equation  of  the  tan- 
gent in  terms  of  its  slope  only.  This  can  be  found  for 
each  of  the  conies  just  as  it  was  found  for  the  circle  in 
Art.  58. 

We  shall  give  the  outline  of  the  work  for  the  ellipse. 
Starting  the  solution  of  the  equation  of  any  line,  y=  lx  +  /3, 
with  the  equation  of  the  ellipse,  b2x2  -f-  a2y2  —  a2b2,  we  have, 
for  obtaining  the  abscissas  of  the  points  of  intersection, 
the  equation 

(52  +  a2?2)  3,2  +  9  aH$x  +  a2  (/32  -  b2)  =  0. 

There  will  be  two  solutions  of  this  equation,  and  hence 

two  points  where  the  line  cuts  the  ellipse.     But  if  (see 

Art.  8) 

4  aH2^  =  4  (b2  +  a2l2~)Qa2P  -  a2b2), 


or  /3  =  ±Va2Z2  +  62, 

the  two  solutions  of  this  equation  are  equal,  the  two  points 
of  intersection  of  the  line  with  the  ellipse  have  become 
coincident,  and  the  line  is  tangent  to  the  ellipse. 


132  ANALYTIC   GEOMETRY  [Ch.  X,  §  77 

The  equation  of  a  tangent  having  a  given  slope  I  is 

therefore  ____. 

y  =  lx±  VaH2  +  V2.  [57] 

Let  the  student  show  that  the  equation  of  the  tangent 
having  a  given  slope  I  is 

for  the  hyperbola,  b2x2  ~  a2y2  =  a2b2, 

y  =  lx±  y/aH2  -  b2,  [58] 

for  the  parabola,  y2=  2 mx, 

y  =  i*  +  fv  [59] 

PROBLEMS 

1.  Find  the  equations  of  the  tangents  to  the  ellipse 
4 x2  -f  y2  =  4  which  are  parallel  to  the  line  2a?  —  4^  +  5  =  0. 

2.  Find  the  equation  of  the  normal  to  the  parabola  y2  =  Sx, 
which  is  parallel  to  the  line  2  x  -f  3  y  =  10. 

3.  Find  the  equations  of  the  tangents  to  the  ellipse 
x2  +  2  y2  —  x  +  y  =  0,  which  are  perpendicular  to  the  line 
x  —  5  y  =  6. 

4.  Find  the  condition  which  must  be  satisfied  if  the  line 

x2     v2 
y  =  lx  +  fl  is  tangent  to  the  hyperbola  —  —  *-  =  —  1. 

5.  Find  the  points  on  each  of  the  conies  where  the  tangents 
are  equally  inclined  to  the  axes.  For  what  case  is  the  solution 
impossible  ? 

6.  Find  the  points  on  the  ellipse  and  hyperbola  where  the 
tangents  are  parallel  to  the  line  joining  the  positive  extremi- 
ties of  the  axes. 

x        1/ 

7.  Show  that  the  line  -  +  \  =  1  is  tangent  to 


a      p 

X2       V2 

(a)  the  ellipse         —  +  fr  =  1,   if 
or      b~ 

«2  +  ^  =  l. 

«2  +  /?2        ' 

o             2 

(b)  the  hyperbola  —  —  ^  =  1,   if 

a2  b2  1 . 
a2      y32 

(c)   the  parabola    y2  =  2  mx,       if 

ma  +  2/32  =  0. 

Sm  X,  §  78 j  TANGENTS  133 

8.  Find  the  equations  of  the  common  tangents  to  the  ellipse 
ar  +  9  y2  =  9,  and  the  circle  x2  +  y2  =  4. 

Note.  —  Write  the  equation  of  the  tangent  to  each  curve  in  the  slope 
form  and  determine  the  value  of  I  which  will  make  the  two  equations 
identical. 

9.  Find  the  equations  of  the  common  tangents  to  the 
ellipse  4  ar  +  9  y2  =  .'56,  and  the  hyperbola  x2  —  y2  =  16.  Show 
that  there  would  be  no  common  tangent,  if  the  second  member 
of  the  equation  of  the  hyperbola  had  any  value  less  than  9. 
Why  ?     Construct  the  figure. 

10.  Find  the  equations  of  the  common  tangents  to  the  circle 
ar  +  y2  =  9,  and  the  parabola  y2  =  Sx.  How  many  solutions 
are  there  ?     Why  ?     Construct  the  figure. 

11.  Show  that  two  tangents  can  be  drawn  to  any  conic  from 
an  exterior  point. 

12.  Through  any  given  point  how  many  normals  can  be 
drawn  to  (a)  an  ellipse,  (b)  a  parabola  ? 

13.  Obtain  the  equation  of  the  tangent  at  the  point  P1  of 
the  parabola  y2  =  2  mx,  by  determining  I  in  the  slope  form  of 
the  equation  of  the  tangent  in  terms  of  xx  and  yx. 

78.  Theorems  concerning  tangents  and  normals.  —  1.  The 
tangent  and  normal  at  any  point  of  an  ellipse  bisect  the 
exterior  and  interior  angles  respectively  between  the  focal 
radii  drawn  to  the  point  of  contact. 

Let  P^T  and  PXN  be  the  tangent  and  normal  to  the 

ellipse  —--\-^-=\  at  the  point  Pv     We  wish  to  show  that 
a2      b2 

PXT  bisects  the  angle  FPXK,  and  that  PXN  bisects  the 
angle  F'P^F. 

It  is  a  well-known  theorem  of  elementary  geometry  that 
the  bisector  of  an  interior  angle  of  a  triangle  divides  the 
opposite  side  into  segments  which  are  proportional  to  the 


134 


ANALYTIC   GEOMETRY 


[Ch.  X,  §  78 


adjacent  sides  of  the  triangle.     The  converse  theorem  is 
also  true. 

It  is  therefore  sufficient  to  show  that 

F'PX  =  F'JV 
FPX       JVF' 

The  equation  of  the  normal  PXN  is 

ahj^x  -  b\y  =  O2  -  b2)xlyv 


Fig.  71. 

a2  —  b2 
Its  intercept  CN  on  the  X-axis  is —  xv  or,  since  in 

a2  —  b2  a 

the  ellipse —  =  e2, 

a2 

CJSr=e2xv 
Also,  FfO=CF=ae. 

Hence  F,N=F,C+CN=  ae  +  e2xx  =  e(a  +  exx), 
and  NF=CF    -  CN=  ae  -  e2xx  =  e(a  -  ex{). 

But  (by  [45]) 

F,P1  =  a  -f-  exv  and   FPX  =  a  —  &rr 


Ch.  X,  §  78] 


TANGENTS 


135 


Hence 


and  the  normal  bisects  the  angle 


F'P,  =  F'N 

FPX       NF' 

F' PXF.     Since  the  tangent  is  perpendicular  to  the  normal, 
it  bisects  the  supplementary  angle  FPXK. 

Note.  —  It  is  upon  this  principle  that  whispering  galleries  are  con- 
structed. If  the  whole  or  part  of  the  sides  of  a  room  is  a  surface  formed 
by  revolving  an  ellipse  about  its  major  axis,  all  waves  of  sound,  light,  or 
heat  starting  from  one  focus  and  striking  this  surface  will  be  reflected  to 
the  other  focus. 

2.  In  an  hyperbola  the  tangent  and  normal  at  any  point 
bisect  the  interior  and  exterior  angles  respectively  between  the 
focal  radii. 

3.  If  an  ellipse  and  hyperbola  are  confocal  (or  have  the 
same  foci),  they  intersect  orthogonally  (or  at  right  angles). 


Fig.  72. 


Since  the  direction  of  a  curve  at  any  point  is  along  the 
tangent  at  that  point,  two  curves  intersect  orthogonally, 
if  their  tangents  at  the  point  of  intersection  are  perpen- 
dicular to  each  other.  This  is  evidently  the  case  here, 
since  the  tangent  to  the  ellipse  bisects  the  exterior  angle 


136 


ANALYTIC    GEOMETRY 


[Ch.  X,  §  78 


between  the  focal  radii,  and  the  tangent  to  the  hvperbola 
bisects  the  interior  angle.  The  curves  therefore  intersect 
orthogonally. 

4.  The  tangent  at  any  point  of  a  parabola  makes  equal 
angles  with  the  focal  radius  drawn  to  the  point  of  contact, 
and  with  the  axis  of  the  curve. 

In  the  parabola  y2  =  2  mx,  let  PXT  and  PXN  be  the  tan- 
gent and  normal  at  Pv     Join  PXF  and  draw  P^K  parallel 


Fig.  73. 


to  OX.     We  wish  to  prove  that  the  angles   TPXF  and 
FTPX  are  equal. 

If  we  let  the  abscissa  of  Px  be  xv  its  ordinate  will  be 
±  V2  mxv  since  Px  is  a  point  on  the  parabola.  Then  the 
equation  of  the  tangent  at  Px  is 

±  ^/2mx1  •  y  =  m  (x  +  x^) . 

If  in  this  equation  we  let  y  =  0,  we  find  the  intercept 
OT  to  be  —x,. 
Hence 


TO 


m 


xv  and    TF=x1  +  ^. 


But         FPX  =  \  (xx  -  |Y  +  2  mx1  =  xx  + 


Cii.  X,  §  78] 


TANGENTS 


137 


Hence   TF=FPV  and  the  angles  TPXF  and  FTP1  are 

equal.     What  other  angles  are  also  equal  in  the  figure  ? 

Note.  —  Parabolic  reflectors  depend  on  this  principle.  If  a  surface  is 
formed  by  revolving  a  parabola  about  its  axis,  all  waves  of  light,  etc., 
which  start  from  the  focus  will  be  reflected  in  lines  parallel  to  the  axis  of 
the  parabola. 

5.  Two  parabolas  which  have  the  same  focus  and  axis, 
but  which  are  turned  in  opposite  directions,  cut  each  other 
orthogonally. 

6.  The  chord  of  contact  of  tangents  to  a  parabola  from 
any  point  on  the  directrix  passes  through  the  focus. 


Fig.  74. 


The  coordinates  of  any  point  L  on  the  directrix  may 
be  represented  by  (  —  — ,  yA.  The  equation  of  PXPV  the 
chord  of  contact  of  tangents  from  this  point,  is 


yxy  =  mx 


m? 


y  = 

lx  + 

m 
21 

X 

ml 

y  = 

~T 

~  ~2 

138  ANALYTIC   GEOMETRY  [Ch.  X,  §  78 

Since  the  coordinates  of  the  focus  (— ,  0  J  satisfy  this 

equation,  the  chord  of   contact  must   pass  through   the 
focus. 

Let  the  student  prove  the  converse  theorem,  viz. :  Tan- 
gents at  the  ends  of  any  focal  chord  meet  on  the  directrix. 

7.  Prove  that  the  same  theorems  hold  for  the  ellipse 
and  hyperbola. 

8.  Any  two  -perpendicular  tangents  to  the  parabola  meet 
on  the  directrix. 

Two  perpendicular  tangents  may  be  represented  by  the 
equations 

and 

By  solving  these  equations  simultaneously,  the  point  of 
intersection  of  the  two  tangents  is  found  to  be 

r_m     m(\  -  P)-| 

which  is  evidently  a  point  on  the  directrix. 

Let  the  student  prove  the  converse  theorem,  viz. :  Two 
tangents  to  a  parabola  from  any  point  on  the  directrix  are 
perpendicular  to  each  other. 

9.  Tangents  to  a  parabola  at  the  ends  of  any  focal  chord 
are  perpendicular  to  each  other. 

10.  Show  that  theorems  8  and  9  do  not  hold  for  cen- 
tral conies,  but  that  perpendicular  tangents  (a)  to  an 
ellipse  meet  on  the  circle  x2  +  y2  =  a2  -j-  b2 ;  (£>)  to  an  hyper- 
bola meet  on  the  circle  x2  +  y2  =  a2  —  b2.  (See  Chap.  14, 
Prob.  1.) 

11.  The  line  joining  any  point  on  the  directrix  of  a  pa- 


Ch.  X,  §  78] 


TANGENTS 


139 


rabola  to  the  focus  is  perpendicular  to  the  chord  of  contact 
of  tangents  from  the  point. 

Take  the  coordinates  of  the  point  L  (Fig.  74)  on  the 
directrix  as  (  —  — ,  yx  ),  and  show  that  the  line  LF  which 
joins  this  point  to  the  focus  is  perpendicular  to  the  chord 


mx 


m- 


of  contact  yxy 

12.  Prove  the  same  theorem  for  the  central  conies. 

13.  The  two  tangents  which  may  be  drawn  from  an 
exterior  point  to  any  conic  subtend  equal  angles  at  the  focus. 

14.  In  the  parabola  the  perpendicular  from  the  focus  on 
any  tangent  meets  it  on  the  tangent  at  the  vertex ;  the  per- 
pendicular meets  the  directrix  on  the  line  through  the  point 
parallel  to  the  axis  of  the  parabola. 

The  equation  of  the 
tangent  at  any  point  (xv 

yd  is 

yxy  =  mx  4-  mxv 
The  equation  of  a  per- 
pendicular to  the  tangent 
through  the  focus  is 

yxx  +  my  =  ^1- 

The  coordinates  of  the 
point  of  intersection  of 
these  two  lines  are 


(».  t> 


Fig.  75. 


They  therefore  meet  on  the  F'-axis,  which  is  the  tan- 
gent at  the  vertex. 

Let  the  student  prove  the  second  part  of  the  theorem, 


140  ANALYTIC    GEOMETRY  [Ch.  X,  §  78 

15.  Show  that  theorem  14  does  not  hold  for  central 
conies,  but  that  the  perpendiculars  from  the  foci  of  a  cen- 
tral conic  on  any  tangent  meet  the  tangent  on  the  circle 
x*  +  y*=  a2.     (See  Chap.  14,  Prob.  5.) 

16.  The  perpendicular  from  a  focus  on  any  tangent  to 
a  central  conic  meets  the  corresponding  directrix  on  the  line 
joining  the  centre  to  the  point  of  contact  of  the  tangent. 

17.  In  any  conic,  tangents  at  the  ends  of  the  latus  rectum 
meet  the  X-axis  on  the  directrix. 

18.  The  tangent  at  any  point  of  the  parabola  meets  the 
directrix  and  latus  rectum  produced  at  points  equally  dis- 
ta n  t  from  the  focus. 

19.  The  product  of  the  perpendiculars  from  the  foci 
of  a  central  conic  on  any  tangent  is  constant  and  equal 
to  P. 

20.  The  semi-minor  axis  b  of  a  central  conic  is  a  mean 
proportional  between  a  normal  and  the  distance  from  the 
centre  to  the  corresponding  tangent. 

21.  The  tangents  at  any  point  of  an  ellipse  and  the  cor- 
responding point  on  the  auxiliary  circle  pass  through  the 
same  point  on  the  axis. 

PROBLEMS 

1.  Show  that,  if  the  point  of  contact  of  a  tangent  to  an 
hyperbola  moves  off  along  the  curve,  the  tangent  approaches 
the  asymptote  as  its  limiting  position. 

2.  Find  the  equations  of  the  tangents  to  the  hyperbola 
which  pass  through  the  centre.  (Use  slope  form  of  the  equa- 
tion of  the  tangent.) 

3.  Show  that  the  portion  of  any  tangent  to  an  hyperbola 
included  between  the  asymptotes  is  bisected  at  the  point  of 
contact. 


Ch.  X.  §  78  ]  TANGENTS  141 

4  Show  that  the  area  of  the  triangle  formed  by  any  tan- 
gent to  an  hyperbola  and  the  asymptotes  is  constant. 

5.  Through  any  point  of  an  hyperbola  parallels  to  the 
asymptotes  are  drawn.  Show  that  the  area  of  the  parallelo- 
gram formed  by  these  lines  and  the  asymptotes  is  constant. 

6.  Show  that  the  normal  at  one  extremity  of  the  latus 
rectum  of  the  parabola  is  parallel  to  the  tangent  at  the  other 
extremity  of  the  latus  rectum. 

7.  Obtain  the  equation  of  the  parabola  referred  to  tangents 
at  the  ends  of  the  latus  rectum  as  coordinate  axes. 

8.  Show  that  the  distances  of  the  vertex  and  focus  of  a 
parabola  from  the  tangent  at  one  end  of  the  latus  rectum  are 
in  the  ratio  of  1 :  2. 

9.  Show  that  the  directrix  of  a  parabola  is  tangent  to  the 
circle  described  on  any  focal  chord  as  a  diameter. 

10.  Show  that  the  tangent  at  the  vertex  of  a  parabola  is 
tangent  to  the  circle  described  on  any  focal  radius  of  the 
parabola  as  a  diameter. 

11.  The  tangent  and  normal  at  a  point  of  the  ellipse  form 
an  isosceles  triangle  with  the  X-axis.  Find  the  coordinates  of 
the  point. 

12.  Prove  that  the  angle  between  two  tangents  to  a  parabola 
is  one-half  of  the  angle  between  the  focal  chords  drawn  to  the 
points  of  contact. 

13.  Show  that  the  length  of  a  normal  in  an  equilateral 
hyperbola  is  equal  to  the  distance  of  the  point  of  contact  from 
the  centre. 

14.  Find  the  points  on  the  conjugate  axis  of  an  hyperbola 
from  which  tangents  to  the  hyperbola  are  perpendicular  to 
each  other. 


CHAPTER    XI 


DIAMETERS 


79.  The  diameter  of  a  conic  may  be  denned  as  the  locus 
of  the  middle  points  of  a  system  of  parallel  chords.  The 
method  of  finding  this  locus  has  already  been  illustrated 
for  the  circle  on  page  94. 

We  shall  repeat  the  work  for  the  ellipse  --  -+-  ^  =  1. 

Let  y  =  lxx  -f-  /3  be  the  equation  of  any  one  of  the  parallel 
chords ;  let  (xv  y^)  and  (z2,  y2)  be  the  coordinates  of  the 


Fig.  76. 


points  where  it  cuts  the  ellipse,  and  (V,  yr%)  the  coordinates 
of  the  point  midway  between  these  two  points. 

142 


Ch.  XI,  §  79] 


DIAMETERS 


143 


Starting  the  solution  of  the  two  equations,  y  —  lxx-\-^ 
and  b2x2  +  a2y2  =  a2b2,  we  have 

(J)2  +  a2!2)  x2  +  2  a\px  +  a2  (/&  -  b2)  =  0, 

the  two  roots  of  which  must  be  x1  and  xT 


But 

Hence 


xf=x1±x. 


x>= *M- 


b2  +  a\2 


[By  Art.  8] 


Since  (V,  y1)  is  on  the  line  y  =  lxx  4-  A  yr  may  be  found 
by  substituting  the  value  of  xf  in  that  equation.  This 
gives 


Hence 


y 


cfil, 

-  *2' 


or,  dropping   primes,  we   have   as   the   equation   of  the 
diameter, 

b2x  +  a-hy  =  0.  [60] 


Fig.  78. 


144  ANALYTIC   GEOMETRY  [Ch.  XI,  §  80 

Let  the  student  show  that  the  equation  of  a  diameter  oi 

^2  2 

the  hyperbola,    ^  —  %-  =  1,  is    b2x  -  a?hy  =  0,     T611 
a2,      ¥ 

the  parabola,       y2'=2mx,  is    l^y-m.  [62] 

The  form  of  the  equation  of  a  diameter  of  an  ellipse  or 
hyperbola  shoivs  that  it  passes  through  the  centre  of  the  conic, 
and  that  it  therefore  conforms  to  the  ordinary  definition 
of  a  diameter.  But  all  the  diameters  of  a  parabola  are 
seen  to  be  parallel  to  the  axis. 

Since  any  value  may  be  given  to  lv  all  lines  through 
the  centre  of  an  ellipse  or  hyperbola  and  all  lines  parallel 
to  the  axis  of  a  parabola  are  diameters. 

Let  the  student  obtain  the  equation  of  the  diameter  of 
each  conic  by  the  following  method :  Transform  to  polar 
coordinates,  with  the  middle  point  (V,  y'j  of  any  one  of 
the  parallel  chords  as  origin.  If  tan-1  lx  is  substituted 
for  6  in  this  equation  of  the  conic,  the  resulting  values  of 
p  should  be  equal  in  magnitude,  but  have  opposite  signs. 
The  necessaiy  condition  for  this  will  be  an  equation  be- 
tween x\  y\  and  lv  which  will  therefore  be  the  equation 
of  the  diameter. 

80.  Conjugate  diameters. — If  we  let  l2  be  the  slope 
of  the  diameter  which  bisects  a  system  of  chords  of  slope 
lv  we  see  that  for  the  ellipse 

7  b2  7  7  & 

and  for  the  hyperbola 

7        b2  77       b* 


Ch.  XI,  §  80] 


DIAMETERS 


145 


Since  in  these  expressions  lx  and  l2  are  interchangeable, 
it  is  evident  that,  if  we  started  out  with  a  system  of 
chords  of  slope  Z2,  the  corresponding  diameter  would  have 
lx  for  its  slope.  Hence  the  two  diameters  which  have  Zx 
and  l2  for  their  slopes  are  so  related  to  each  other  that  each 
bisects  all  chords  parallel  to  the  other.  Such  diameters 
are  said  to  be  conjugate  to  eacli  other.     Their  equations 


are 


y  =  lxx,  and  y  =  l2x, 


where  Zx  and  l2  are  connected  by  the  relations  given 
above.  In  the  ellipse,  lx  and  l2  have  opposite  signs,  and 
the  diameters  must  pass  through  different  quadrants. 
But  in  the  hyperbola,  lx  and  l2  have  the  same  sign,  and 
the  diameters  must  pass  through  the  same  quadrant.  In 
either  case,  as  lx  decreases  in  numerical  value,  l2  increases, 
and  as  one  diameter  approaches  the  major  axis,  the  other 
will  approach  the  minor  axis  from  one  side  or  the  other. 
The  limiting  case  is  seen  to  be  the  two  axes.  They  con- 
form to  the  definition 
of  conjugate  diameters, 
since  every  line  paral- 
lel to  one  is  bisected 
by  the  other.  They 
are  the  only  conjugate 
diameters  which  are 
perpendicular  to  each 
other. 

If  in  the  ellipse  one 
diameter,  PXKV   starts 
coincident  with  the  major  axis  and  revolves  in  the  posi- 
tive direction,   its   conjugate   diameter,   P2K2,   will   start 


Fig.  79. 


146 


ANALYTIC   GEOMETRY 


[Ch.  XI,  §  80 


coincident  with  the  minor  axis  and  also  revolve  in  the 
positive  direction,  since  we  have  seen  that  the  two  must 
pass  through  different  quadrants. 

Let  the  student  show  that  the  angle  PXCPV  in  which 
the  minor  axis  lies,  will  always  be  obtuse. 

If,  in  the  hyperbola,  J>1K1  starts  coincident  with  the 
major   axis    and   revolves   in  the   positive   direction,  its 


conjugate  diameter,  P2K2,  will  start  coincident  with  the 

minor  axis  and  revolve  in  the  negative  direction.      For 

the  two  diameters  must  remain  in  the  same  quadrant.    Since 

b2  b 

the  product  of  the  two  slopes  is  — ,  if  L  is  less  than  -,  /„ 
,  a*  a 

must  be  greater  than  -,  and  the  two  diameters  must  there- 
a 

fore   rernain   on   opposite    sides  of  the    asymptote.     As   lx 

approaches  -,  L  also  approaches  -,  and   the   asymptote 
a  a 

is  therefore  the  limiting  position  of  both  diameters.     It 


Ch.  XI,  §  81]  DIAMETERS  147 

is  evident  that  only  one  of  two  conjugate  diameters  can  cut 
the  hyperbola.  But  in  speaking  of  the  length  of  the  other 
diameter,  we  shall  mean  the  distance,  P2iT2,  between  the 
points  where  it  cuts  the  conjugate  hyperbola. 

81.    Equation  of  conjugate  diameter. — If  one  diameter 

is  n'iven  in  any  way,  as  by  means  of  its  slope  or  the  coor- 
dinates of  its  extremity,  it  is  easy  to  determine  the  con- 
jugate diameter.  For  it  is  only  necessary  to  write  the 
equation  of  a  line  through  the  centre  whose  slope  bears 
the  required  relation  to  the  slope  of  the  given  line.  If 
the  coordinates  of  Pl  (Fig.  79)  are  (xv  yx),  the  equa- 
tion of  CPl  is  xxy  —  yxx  =  0,  and  lx  =  — .  Then  for  the 
ellipse,  * 

I  =       h2  =      ^ , 
2  a\  a^yl 

and  the  equation  of  the  conjugate  diameter  P2^2  *s 

9^  +  mii  =  0.  T631 

a2        b~  L     J 

The  solution  of  this  equation  with  the  equation  of  the 

ellipse  gives  for  the  coordinates  of  P9  ( f*i    — l ),  and 

(ay       bx  \  "  ^        "        a  ' 
for  the  coordinates  of  K0     -f1 l  • 

2  \  b         a  J 

Let  the  student  show  that  in  the  hyperbola  the  equa- 
tion of   a  diameter  conjugate   to  the  diameter  through 

(a\,  vO  is 

£!£_  2i2  =  0,  [64] 

a2        b-  L      J 

and  that  the  coordinates  of  its  extremities  are 
fay,    bx,\        if     ay,  bx,\ 


148  ANALYTIC   GEOMETRY  [Ch.  XI,  §  82 

PROBLEMS 

1.  Find  the  equations  of  a  pair  of  conjugate  diameters  of 
the  hyperbola  x2  —  8  y2  =  96,  one  of  which  bisects  the  chord 
whose  equation  is  3  x  —  8  y  =  10. 

2.  Find  the  equation  of  the  diameter  of  the  parabola 
y2  =  6x,  which  bisects  all  chords  parallel  to  the  line  x+3y  =  8. 

3.  Find  the  equation  of  a  diameter  of  the  ellipse 

£x2  +  9y2  =  36, 
if  one  end  of  its  conjugate  diameter  is  (f  V3,  1). 

4.  What  is  the  equation  of  the  chord  of  the  ellipse 
9  x2  +  36  f  =  324,  which  is  bisected  by  the  point  (4,  2)  ? 

5.  Find  the  equation  of  a  chord  of  the  ellipse 

13ar9  +  ll?/2  =  113 

through  the  point  (1,  3),  which  is  bisected  by  the  diameter 
2y  =  3x. 

6.  A  diameter  of  the  ellipse  15 1/2  +  4  x2  =  60  is  drawn 
through  the  point  (1,  f).  Find  the  equation  of  the  conjugate 
diameter  and  its  points  of  intersection  with  the  ellipse. 

7.  Find  the  length  of  the  diameter  of  the  hyperbola 
9  x2  —  4  y2  =  36,  which  is  conjugate  to  the  diameter  y  =  3  x. 

8.  What  is  the  equation  of  the  chord  of  the  parabola 
y2  =  6  x,  which  is  bisected  by  the  point  (4,  3)  ? 

9.  A  diameter  of  the  hyperbola  4  x2  —  16  y2  =  25  passes 
through  the  point  (1,  —  2).  Find  its  extremities  and  the 
extremities  of  its  conjugate  diameter. 

10.  What  is  the  relation  between  the  slopes  of  the  con- 
jugate diameters  of  the  equilateral  hyperbola  xy  =  k? 

82.  Theorems  concerning  diameters. — 1.  In  the  ellipse 
the  sum  of  the  squares  of  any  tivo  conjugate  semi-diameters 
is  equal  to  the  sum  of  the  squares  of  the  semi-axes. 


Ch.  XI,  §  82] 


diameters 


149 


In  Fig.  81,  let  CPt=  a!  and  0P2=  b'.     Then  (by  [1]) 


and 


.12  —  >.  2 


bz  a2 


Fig.  81. 


Adding,  a'2  +  b'2  =  (a2  +  62)  ^  +  (a2  +  &2) 

But  since  Px  is  on  the  ellipse, 

%+^r  =  l,    and    a'2  +  b'2  =  a2  +  b2. 

az       bz 


2.  In  the  hyperbola  the  difference  of  the  squares  of  any 
two  conjugate  semi- diameters  is  equal  to  the  difference  of  the 
squares  of  the  semi-axes. 

3.  The  product  of  the  focal  distances  of  any  point  on  a 
central  conic  is  equal  to  the  square  of  the  semi-diameter 
conjugate  to  the  diameter  through  the  point. 


150  ANALYTIC   GEOMETRY  [Ch.  XI,  §  82 

Since  the  focal  distances  of  the  point  (xv  y^)  have  been 
shown  (Art.  70)  to  be  a  -f-  ex1  and  a  —  exv  it  is  necessary 
to  show  that  b'2  =  a2  —  e2xx2.     From  theorem  1  we  have 

b2    +    a2' 


But  since  Px  is  a  point  on  the  ellipse, 

W  _ 

J2     --       "-I" 


tfy*  =  a2b2-b2x2,    or  ^-  =  a2 


b2x  2 
Hence       5'2  =  a2  —  x?  -\ ^-, 

1  a2 


ai-f'sL^)^,  or  (by  [39]), 


=  a2  —  A-,2. 


4.  Prove  the  same  theorem  for  the  hyperbola. 

5.  TAe  tangents  at  the  ends  of  a  diameter  of  a  central 
conic  are  parallel  to  the  conjugate  diameter. 

In  the  ellipse  the  equation  of  the  tangent  at  P1  is 

a2  ^  b2  ~~ 

This  is  seen  at  once  to  be  parallel  to  the  conjugate 

diameter    -^  +  ¥j¥-  =  0.      In  the  same  way  the  tangent 

at  P2  can  be  shown  to  be  parallel  to  PVKV 

This  theorem  appears  also  from  the  fact  that  the  tan- 
gents are  special  cases  of  the  system  of  parallel  chords. 

6.  The  tangent  at  the  end  of  a  diameter  of  the  parabola 
is  parallel  to  the  system  of  chords  which  the  diameter  bisects. 

7.  The  area  of  the  parallelogram  formed  by  tangents  at 
the  ends  of  conjugate  diameters  of  a  central  conic  equals  the 
area  of  the  rectangle  on  the  principal  axes. 


Oh.  XI,  §  82] 


DIAMETERS 


151 


Let  ABED  be  the  parallelogram  formed  by  the  tangents 
at  the  ends  of  the  conjugate  diameters  PXKX  and  P2K2. 


B^^ 

Y 

Ji 

sl — - — y^       \ 

^\T^ 

\                            T  \ 

\     y^^^A 

D 

z^~~Ks 

Fig.  82. 


The  sides  of  the  parallelogram  are  evidently  2  a1  and  2  V '. 
From  (7  drop  a  perpendicular  (7M"  on  AB.  Its  length  is 
the  distance  from  the  origin  to  the  line 

b2xxx  +  a2^?/  =  a2b2,  or  (by  [17]) 

a%2 


CM= 


■Vb%2  +  a4^2 
But  the  area  of  the  parallelogram 


=  2  OMx  AB=2  CMxP2K2, 


2  a*b* 


-J 


152  ANALYTIC   GEOMETRY  [Ch.  XI,  § 

Let  the  student  give  the  proof  for  the  hyperbola. 

r 


8.  In  the  hyperbola  the  parallelogram  formed  by  the 
tangents  at  the  ends  of  conjugate  diameters  has  its  vertices 
on  the  asymptotes. 

9.  In  the  hyperbola,  the  line  joining  the  ends  of  conju- 
gate diameters  is  parallel  to  one  asymptote  and  is  bisected 
by  the  other. 

10.    Shoiv  that  the  angle  between  two  conjugate  diameters 
i  ab 


sin 


']' 


a'b 


11.  Conjugate  diameters  of  the  rectangular  hyperbola  are 
equal. 

12.  The  ellipse  has  a  pair  of  equal  conjugate  diameters 
which  coincide  with  the  asymptotes  of  the  hyperbola,  which 
has  the  same  axes  as  the  ellipse. 


Ch.  XI,  §  82]  DIAMETERS  153 

PROBLEMS 

1.  Prove  that  conjugate  diameters  of  an  equilateral  hyper- 
bola are  equally  inclined  to  the  asymptotes. 

2.  Prove  that  any  two  perpendicular  diameters  of  an  equi- 
lateral hyperbola  are  equal. 

3.  Prove  that  the  straight  lines  drawn  from  any  point  in 
an  equilateral  hyperbola  to  the  extremities  of  any  diameter 
are  inclined  at  equal  angles  to  the  asymptotes. 

4.  Prove  that  the  points  on  either  the  major  or  minor 
auxiliary  circle,  which  correspond  to  the  extremities  of  a 
pair  of  conjugate  diameters,  subtend  a  right  angle  at  the 
centre  of  the  ellipse. 

5.  Prove  that  chords  drawn  from  any  point  of  a  central 
conic  to  the  extremities  of  a  diameter  (called  supplemental 
chords)  are  always  parallel  to  a  pair  of  conjugate  diameters. 

6.  If  Pi  and  P2  are  the  extremities  of  a  pair  of  conjugate 
diameters  of  a  central  conic,  prove  that  the  normals  at  Px  and 
P2,  and  the  perpendicular  from  the  centre  on  P2P2  meet  in  a 
point. 

7.  If  a  perpendicular  is  drawn  from  the  focus  of  a  central 
conic  to  a  diameter,  show  that  it  meets  the  conjugate  diameter 
on  the  corresponding  directrix. 

8.  Tangents  at  the  extremities  of  a  pair  of  conjugate  diam- 
eters of  an  ellipse  form  a  parallelogram  (Fig.  82).  Show  that 
the  diagonals  of  this  parallelogram  are  also  a  pair  of  conjugate 
diameters. 


CHAPTER   XII 
POLES    AND  POLARS 

83.    Harmonic  division.  —  In  Art.  14  we  have  said  that 

four  points  A,  B,  C,  and  D  on  a  line  form  a  harmonic 

.,  AB         AD       A  ,,    , 
4 b     c  p    range,  if  —  =  -  — ,  and  that 

FlG-  84-  the  pairs  of  points  A  and  C,  and 

B  and  D  are  then  called  conjugate  harmonic  points.  From 
this  definition  it  is  easily  seen  that  if  we  keep  A  and  C 
fixed  and  allow  B  and  D  to  move,  as  B  approaches  C,  B 
will  also  approach  C,  and  as  B  approaches  the  middle 
point  of  AC,  B  will  recede  indefinitely.  When  B  coin- 
cides with  the  middle  point  of  AC,  it  has  no  conjugate 
harmonic  point.  When  B  moves  from  the  centre  toward 
A,  B  comes  in  from  the  left  toward  A. 

It  is  desirable  for  our  purposes  to  express  the  relation 
between  these  points  in  terms  of  distances  from  A  only. 
From  the  definition,  AB  x  CD  =  AD  x  BC.  Substituting 
CD  =  AD -AC  and  BC=  A  C  -  AB,  this  becomes 

ABxAD-ABx  AC=  AD  x  AC -  AD  x  AB, 

2AB  xAD 


or  AC  = 


AB  +  AD 


Let  the  student  show  that  BD  =  -— — —- — .     Connect 

BA  +  BC 

these  results  with  harmonic  progression  in  algebra  by  show- 
ing that  A  C  is  a  harmonic  mean  between  AB  and  AD, 

154 


Ch.  XII,  §  84] 


POLES   AND   POLARS 


155 


84.  Polar  of  a  point.  —  The  polar  of  a  point  with  respect 
to  any  conic  is  defined  as  the  locus  of  points  which  divide 
harmonically  secants  through  the  fixed  point. 

The  methods  of  finding  this  locus  are  the  same  for  all 
the  conies.  In  problem  31,  Chapter  VIII,  the  student 
has  been  asked  to  find  it  for  the  circle  by  the  aid  of  polar 


— P, 


Fig.  85. 

coordinates.  The  same  method  might  be  employed  here, 
but  it  is  thought  best  to  use  a  very  similar  one,  into 
which,  however,  polar  coordinates  do  not  enter. 

We  shall  find  the  polar  of  the  point  Px  with  respect  to 
the  ellipse  ^  +  ay  =  a^ 

Transform  to  the  point  Px  as  origin  by  the  aid  of  the 


equations 


x  =  xf  +  X, 


—  />/' 


y  +  Vv 


156  ANALYTIC   GEOMETRY  [Ch.  XII,  §  84 

The  equation  of  the  ellipse  becomes 

b2x2  +  a2y2  +  2  b2xxx  +  2  a2yxy  +  b2x2  +  a2?^2  -  a262  =  0. 

Let  any  line,  y  —  Ix,  through  P1  cut  the  ellipse  in  the 
points  P2  and  P3.  We  wish  to  find  the  locus  of  a  point 
P'  on  this  line,  so  situated  that  Pv  P2,  P',  and  P3  form 
a  harmonic  range.  By  the  theorem  of  Art.  14,  Pv  M2, 
M,  and  Mz  will  also  form  an  harmonic  range,  and  hence 

If  we  start  the  solution  of  the  equation  of  the  line, 
y  =  lx,  with  the  equation  of  the  ellipse,  we  have 


(b2  +  a2/2)  x2  +  2 \b2xx  +  aHyJ  x  +  b2x2  +  a2y2  -  a2b2  =  0, 

rom  which  t 
(see  Art.  8) 


from  which  to  determine  the  values  of  x2  and  xs.     Hence 


_  b2x2  +  a2y,2  -  a2b2 
X^-  &  +  <£!* 

A  ,  2(b2xl  +  a2ly1) 

and  x2  +  xs  =  -     V+«2Z2 

Hence  ,^,^4-^-^ 

o1xl  +  «  t^i 

Since  P'  lies  on  the  line  y  =  Ix,  its  coordinates  satisfy 

the  equation,  and  y'  =  lx\   or   l  =  ^-f     Substituting  this 

SB 

value  of  I  in  the  equation  above,  and  reducing,  we  have 
as  the  equation  of  the  locus,  referred  to  the  point  Px  as 
origin, 

b2xxx  +  (Py^y  4-  b2x?  +  a2yx2  —  a2b2  =  0. 


Cii.  XII,  §  85]  POLES  AND   POLARS  157 

When  transformed  back  to  the  original  origin  0  by  the 

aid  of  the  equations  x=x'  —  xx  and  y  =  y'  —  yv  this  equation 

becomes 

b2xix  +  oryx  y=  a2b2,  [65] 

Since  this  equation  is  of  the  first  degree,  we  have 
arrived  at  the  singular  result  that  the  locus  is  a  straight 
line.  It  is  called  the  polar  of  the  point  Pl  with  respect 
to  the  ellipse,  while  the  point  Px  is  called  the  pole  of  the 
line.  The  theory  of  poles  and  polars  is  of  great  impor- 
tance in  some  branches  of  geometry. 

Let  the  student  show  that  the  polar  of  the  point 
(xv  y^)  with  respect  to 

(a)  the  circle,  x2  -f  y2  =  r2,  is  xxx  +  yxy  =  r2,  [Q6] 

(b)  the  hyperbola, 

b2x>  -  a2y2  =  a2b2,  is  b2xxx  -  a2yxy  =  a2b2,  [67] 
(e)  the  parabola,  y2  =  2  mx,  is  yxy  =  mx  +  mx\,  [68] 
(c?)  the  locus  of  the  general  equation, 

Ax2  +  Bxy  +  Cy2  +  Dx  +  Uy  +  F=0, 


is  Axxx  +  —  (yxx  +  xxy)  +  Cyxy 

+  ¥>(x  +  xl)  +  ^(y  +  yl)+F 


[69] 


85.  Position  of  the  polar. — From  what  we  have  said 
about  harmonic  ranges,  it  is  evident  that  the  polar  of  every 
point  outside  the  conic  cuts  the  conic,  and  that  the  polar 
of  every  point  within  the  conic  does  not  cut  it.  The  form 
of  the  equation  shows  that,  when  the  point  is  outside 
the    conic,  the  polar   coincides  with   the    chord   of  contact 


158  ANALYTIC   GEOMETRY  [Ch.  XII,  §  85 

and,  when  the  point  is  on  the  conic,  the  polar  becomes  the 
tangent.  Again,  as  Px  recedes  from  the  conic,  its  con- 
jugate harmonic  point  P'  approaches  the  middle  of  the 
chord,  and  its  polar,  therefore,  approaches  coincidence 
with  a  diameter.  If  the  point  Pl  is  inside  a  central  conic 
and  approaches  the  centre,  the  polar  evidently  recedes 
indefinitely. 

PROBLEMS 

1.  What  is  the  equation  of  the  polar  of 

(a)  (1,  —  2)  with  respect  to  the  conic  of'  +  4  y2  =  16  ? 
(6)  (6,  —  4)  with  respect  to  the  conic  y2  =  4  #  ? 

(c)  (—  3,  2)  with  respect  to  the  conic  5  x2  —  8  y2  =  24  ? 

(d)  (0,  0)       with  respect  to  the  conic 

x2+2xy  +  3y2-4:x-l0  =  0? 

2 .  What  is  the  pole  of  the  line  3  x  —  2  y  =  5  with  respect 
to  the  circle  x2  +  y2  =  25  ? 

Solution.  —  The  polar  of  the  point  (xi,  y{)  with  respect  to  the  circle 
is  X\X  +  y\y  =  25.  We  wish  to  find  the  values  of  Xi  and  yx  which  will 
make  this  line  coincident  with  the  line  3  x  —  2  y  =  5,  or  15  x  —  10  y  =  25. 
They  are  evidently  xi  =  15  and  y±  =  —  10. 

3.  What  is  the  pole  of  the  line  5  x  +  ky  =  7  with  respect 
to  the  ellipse  x2  +  2y2=10? 

4.  What  is  the  pole  of  the  line  x  —  y  =  10  with  respect  to 
the  parabola  y2  =  Sx? 

5.  What  is  the  pole  of  the  line  Ax+By+  C=0  with  respect 
to  the  hyperbola  b2x2  —  a2y2  —  a2b2  ? 

6.  Tangents  are  drawn  to  the  circle  y2  =  10x  —  x2  at  the 
points  where  it  is  cut  by  the  line  y  =  4  x  —  7.  What  is  their 
point  of  intersection  ? 

7.  Through  the  point  (#1?  y{)  a  line  is  drawn  parallel  to  the 
polar  of  the  point  with  respect  to  the  ellipse  b2x2  +  a2y2  =  a2b'2. 
Find  the  coordinates  of  the  pole  of  this  parallel. 


Ch.  XII,  §  86] 


POLES   AND   TOLA  US 


159 


86.  Theorems  concerning  poles  and  polars.  —  1.  If  a 
set  of  points  lie  on  a  line,  their  polars  all  p>a8S  through 
the  pole  of  that  line  ;  and  conversely,  if  a  set  of  lines  pass 
through  a  point,  their  poles  lie  on  the  polar  of  that  point. 

Let  P2  be  the  pole  of  the  line  MN,  and  P1  any  point 
on  MN  We  wish  to  show  that  MS,  the  polar  of  Pv 
will  pass  through  P2.  If  the  coordinates  of  Px  and  P2 
are  (xv  yx)  and  (x2,  y2), 
the  equation  of  PS  is 

b2xxx  +  a2yxy  =  a2b2, 

and  of  MN  is 

b2x2x  +  a2y2y  =  a2b2. 

But  we  know  that  the 
coordinates  of  P1  must 
satisfy  the    equation   of 

MN,  or 

b2x2xx  +  a%y1  =  a2b2. 

Now  this  is  just  the  condition  which  must  be  satisfied, 
if  P2  lies  on  MS.  Hence  P2  lies  on  MS,  and  as  the  point 
Px  moves  along  the  line  MN,  its  polar  will  revolve  about 
P2,  the  pole  of  MN 

Let  the  student  prove  the  converse  theorem,  and  also 
both  theorems  for  the  hyperbola  and  parabola. 

It  follows  from  this  theorem  that  tangents  at  the 
extremities  of  any  chord  through  Px  meet  on  MS.  For 
the  pole  of  every  chord  through  Px  lies  on  MS,  and  we 
have  seen  (Art.  85)  that  tangents  at  the  extremities  of 
a  chord  intersect  at  the  pole  of  that  chord.  From  this 
property  the   polar  may  be  defined  as  the  locus  of   the 


160 


ANALYTIC    GEOMETRY 


[Ch.  XII,  §  86 


intersection    of   tangents   at    the    extremities   of    chords 
through  any  fixed  point. 

This  property  enables  us  to  construct  the  polar  of  any 
point ;    for  any  number  of   points  on  the  polar  may  be 


Fig.  87. 

determined  by  finding  the   intersections   of   tangents  at 
the  extremities  of  chords  through  the  point. 

2.  The  polar  of  any  point  P1  with  respect  to  a  central 
conic  is  parallel  to  the  tangent  at  the  point  where  the  diameter 
through  P1  cuts  the  conic. 

Y 


Fig.  88. 


Ch.  XII,  §  86] 


POLES   AND   POLARS 


161 


Let  the  coordinates  of  the  point  P2  where  CP1  cuts  the 
hyperbola  be  (xv  ya).     Then  the  equation  of  the  tangent 

2   k  b2x2x  —  a2y2y  =  a2b2, 

and  the  equation  of  the  polar  of  Px  is 
b2xxx  —  a2yxy  =  a2b2. 

But  since  Px  and  P2  are  on  the  same  line  through  the 

x       x 
origin,  —  =  — -,  and  these  lines  are  evidently  parallel. 

Let  the  student  prove  the  same  theorem  for  the  ellipse. 

3.  The  polar  of  any  point  Px  with  respect  to  a  parabola  is 
parallel  to  the  tangent  at  the  point  where  a  diameter  through 
P1  cuts  the  parabola. 

Y 


Fig.  89. 


We  may  let  the  coordinates  of  P2  be  (xvy-[)>  Then 
the  equation  of  the  tangent  at  P2  is  yxy  =  mx  +  mxv  and 
the  equation  of  the  polar  of  P1  is  yxy  =  mx  +  mxv  These 
equations  are  seen  at  once    to  represent  parallel   lines. 


162  ANALYTIC   GEOMETRY  [Ch.  XII,  §  86 

These  two  theorems  show  that  the  polar  of  a  point  on  a 
diameter  is  one  of  the  system  of  parallel  chords  bisected 
by  that  diameter. 

4.  If  the  line  joining  the  centre  C  of  any  central  conic  to 
any  point  Px  cuts  the  conic  in  P2  and  the  polar  of  P1  in  P3, 
then  CP1  x  CP3  =  OP*. 

We  shall  give  the  proof  for  the  hyperbola,  using 
Fig.   88. 

The  equation   of  QPX  is  y  =  —  x.     The  coordinates  of 

P2,  where  this  line  cuts  the  hyperbola  are  found  to  be 

ahxl  ~~A  <%1 


and 


-\Jb2x^  —  a2y-f  ^/b2xx2  —  a2y^ 

and  the  coordinates  of  P3,  where  it  cuts  the  polar, 

b2xxx  —  a2yxy  —  a2b2, 

a2b2xx  ,         a2b2y1 

b2xx2  —  a2yx2  b2xx2  —  a2yx2 


Hence  CPX  =  ^x2  +  y2, 


^2~  V    b2x2-a2y2 


_g%^X*  +  y* 

3  "   b2x2  -  a2y2  ' 

From  these  values  we  see  at  once  that 

CPX  x  CP3  =  OP2. 

Let  the  student  prove  the  same  theorem  for  the  ellipse. 
Show  that  in  the  parabola   (Fig.  89)   P2   bisects   the 
line  PXPV 


Ch.  XII,  §  80]  POLES   AND   POLARS  163 

5.  The  line  which  joins  any  point  to  the  centre  of  a  circle 
is  perpendicular  to  the  polar  of  the  point  with  respect  to  the 
circle. 

The  proof  of  this  theorem  appears  at  once  as  soon  as  the 
equations  of  the  lines  are  written.  This  theorem  enables 
us  to  state  theorem  4  for  the  circle  as  follows  : 

6.  The  radius  of  a  circle  is  a  mean  proportional  between 
the  distance  from  the  centre  to  any  point  and  the  distance 
from  the  centre  to  the  polar  of  that  point. 

7.  The  polar  of  the  focus  is  the  directrix  in  (a)  the  ellipse, 
(b)  the  hyperbola,  (c)  the  parabola. 

The  proof  of  this  theorem  appears  at  once  in  each  case 
when  the  coordinates  of  the  focus  are  substituted  in  the 
equation  of  the  polar.  This  theorem  is  evidently  equiva- 
lent to  theorems  6  and  7  on  tangents. 

8.  Any  chord  through  the  focus  of  a  conic  is  perpendicular 
to  the  line  joining  its  pole  ivith  the  focus. 

This  theorem  is  equivalent  to  theorems  11  and  12  on 
tangents  and  is  proved  in  the  same  manner. 

9.  The  line  joining  the  centre  of  a  central  conic  to  any 
point  P1  cuts  the  directrix  in  K.  Show  that  the  line  KF  is 
perpendicular  to  the  polar  of  Pv 

10.  Two  triangles  are  so  related  that  the  vertices  of  the 
first  are  the  poles  of  the  sides  of  the  second,  with  respect 
to  a  conic.  Prove  that  the  vertices  of  the  second  are  also 
poles  of  the  sides  of  the  first. 

Two  such  triangles  are  said  to  be  conjugate  to  each  other. 
If  in  any  triangle  the  vertices  are  the  poles  of  the  opposite 


104  ANALYTIC   GEOMETRY  [Oh.  XII,  §  86 

sides,   the  triangle  and  its  conjugate  coincide,  and  it  is 
called  a  self-conjugate  triangle. 

11.  If  a  line  is  drawn  through  a  point  parallel  to  the  axis 
of  a  parabola,  that  portion  of  it  included  between  the  point 
and  its  polar  is  bisected  by  the  parabola. 

How  does  this  conform  to  the  definition  of  harmonic 
division  ? 

12.  The  two  lines,  which  join  the  focus  of  a  conic  to  any 
point  ctnd  to  the  intersection  of  the  polar  of  that  point  ivith 
the  corresponding  directrix,  are  perpendicular  to  each  other. 

13.  Write  the  equation  of  the  polar  of  a  point  P1  on 
a  diameter  of  a  central  conic.  Let  Px  recede  indefinitely 
along  the  diameter  and  shoiv  that  the  polar  approaches,  as  its 
limiting  position,  the  diameter  conjugate  to  the  given  diameter. 
Show  that  this  would  be  true,  if  the  point  receded  along 
any  line  parallel  to  the  given  diameter.  How  must  this 
theorem  be  stated  for  the  parabola  ? 

PROBLEMS 

1.  Show  that  the  polars  of  the  same  point,  with  respect 
to  two  conjugate  hyperbolas,  are  parallel. 

2.  Show  that  the  four  points,  in  which  any  line  is  cut  by 
the  asymptotes  of  an  hyperbola  and  by  a  pair  of  conjugate 
diameters,  form  a  harmonic  range. 

3.  What  is  the  polar  of  the  focus  of  an  ellipse,  with 
respect  to  the  major  auxiliary  circle  ? 

4.  Obtain  the  equation  of  the  polar  of  the  point  Px  with 
respect  to  the  rectangular  hyperbola  xy  =  k.  What  are  the 
coordinates  of  the  foci  and  the  equations  of  the  directrices 
of  this  hyperbola?  Prove  that  your  results  are  correct  by 
showing  that  the  directrix  is  the  polar  of  the  focus. 


Ch.  XII,  §  86]  POLES   AND  POLARS  165 

5.  Show  that  the  polar  of  one  extremity  of  a  diameter  of 
an  hyperbola,  with  respect  to  its  conjugate  hyperbola,  is  the 
tangent  at  the  other  extremity  of  the  given  diameter. 

6.  If  a  perpendicular  is  let  fall  from  any  point  Px  upon 
its  polar,  prove  that  the  distance  of  the  foot  of  this  per- 
pendicular from  the  focus  is  equal  to  the  distance  of  the 
point  Px  from  the  directrix  of  the  parabola. 

7.  An  ellipse  and  an  hyperbola  have  the  same  transverse 
and  conjugate  axes.  Show  that  the  polar  of  any  point  on 
either  curve,  with  respect  to  the  other,  is  tangent  to  the 
first  curve. 


CHAPTER  XIII 
GENERAL  EQUATION  OF  THE  SECOND  DEGREE 

87.  We  have  seen  that  the  equations  of  all  the  conies 
are  of  the  second  degree.  We  shall  now  prove  that  an 
equation  of  the  second  degree  must  always  represent  a 
conic,  either  in  one  of  the  ordinary  forms  or  in  one  of  the 
limiting  cases,  and  show  how  to  reduce  any  given  equation 
to  the  simplest  equation  of  one  of  these  conic  sections. 

88.  Two  straight  lines.  —  We  have  seen  that  there 
are  certain  equations  of  the  second  degree  which  can  be 
factored,  and  hence  represent  two  straight  lines. 

Let  us  determine  what  condition  must  be  satisfied  by 
the  coefficients  of  the  general  equation, 

(1)  Ax2  +  Bxy  +  Cy2  +  Bx  +  Ey  +  F=  0, 

when  it  can  be  separated  into  two  linear  factors.    Arrang- 
ing it  according  to  the  powers  of  x  and  solving,  we  have 

(2)  x  = 

-  (By  +  i>)  ±  ^/(B2  -  4  AO)y2  +  ( 2  BB  -  ±AE)y  +  B2  -  4  AF. 

2A 

If  the  general  equation  is  to  be  factored  into  two 
linear  factors,  the  quantity  under  the  radical  must  be 
a  perfect  square.     The  condition  for  this  is 

(3)  (2BB-4AB)2-  ±(B2  -  ±  AC)(B2  -  4  AF)  =  0, 

or        (4)  4ACF+BBE-AE2-CB2-FB2  =  Q. 

166 


Ch.  XIII,  §  88]     EQUATION   OF   THE   SECOND   DEGREE  167 

This  is,  then,  the  condition  which  must  be  satisfied  by 
the  coefficients  of  the  general  equation,  when  it  can  be 
separated  into  two  linear  factors.  The  first  member  is 
called  the  discriminant  of  the  equation.  It  is  usually 
represented  by  the  letter  A. 

Note.  —  If  A  =  0,  the  work  will  have  to  be  changed  somewhat,  but  the 
same  form  will  always  be  obtained  for  the  discriminant. 

When  this  condition  is  satisfied,  the  equation  can  always 
be  factored,  but  it  is  not  necessary  that  the  factors  should 
be  real.  For  if  B2 - 4  A  C  is  negative,  from  (3),  B2-  4  AF 
must  also  be  negative,  and  while  the  expression  under  the 
radical  is  a  perfect  square,  its  square  root  will  contain 
imaginary  coefficients.  The  equation  will  in  this  case 
break  up  into  a  pair  of  factors  with  imaginary  coefficients, 
and  we  speak  of  it  as  representing  a  pair  of  imaginary 
lines.  There  will  be,  however,  one  real  point  on  the 
locus ;  for  the  intersection  of  the  two  imaginary  lines  will 
always  be  a  real  point. 

If  B2  —  4  AC  is  positive,  the  factors  represent  real  and 
intersecting  lines. 

If  B2-±AC=0,  2  BD-  ±AE  must  also  reduce  to 
zero,  and  the  quantity  under  the  radical  is  reduced  to 
D2  —  4  AF.     The  lines  are  therefore  parallel.     They  are 

real  and  distinct  if         D2  -  4  AF>  0, 

real  and  coincident  if    D2  —  4  AF  =  0, 

imaginary  if  D2  —  4  AF  <  0. 

PROBLEMS 
1.    Obtain  the  discriminant  by  the  following  method:  Let 
the  two  factors  be  x  -f  bxy  -f-  <h  and  x  -f  b2y  +  c2.     Multiply,  and 
equate  the  coefficients  of  the  product  and  those  of  the  general 


168  ANALYTIC   GEOMETRY  [Ch.  XIII,  §  89 

equation.  This  will  give  five  equations  from  which  b^  cXl  b2, 
and  c2  can  be  eliminated  and  the  condition  in  terms  of  the 
coefficients  obtained. 

2.  Show  that  the  following  equations  represent  straight 
lines,  and  find  the  factors  in  each  case : 

f  —  xy  —  5  x  +  5  y  =  0, 

2x2  +  3xy  +  y9--x-y  =  0, 

x*  +  2xy  +  if  +  2x  +  2y  +  l  =  Q. 

Case  I.     B2-  4  AC^O. 

89.    Removal  of  the  terms  of  the  first  degree.  —  If  the 

discriminant  does  not  vanish,  and  if  the  equation  does 
represent  some  conic,  it  ought  to  be  possible  by  suitable 
transformations,  either  by  changing  the  position  of  the 
origin  or  by  revolving  the  axes,  or  both,  to  reduce  it  to 
one  of  the  well-known  forms. 

Let  us  transform  to  a  new  origin  (#0,  y0)  and  find  the 
values  of  x0  and  y0,  if  any,  which  will  simplify  the  equa- 
tion.    The  general  equation  becomes 

(5)  Ax2  +  Bxy  +  Cy2  +  D'x  +  E' y  +  F'  =  0, 
where     (6)  D'  =  2  Ax0  +  By0  +  D, 

(7)  E<  =  Bx0  +  2  Cy0  +  E, 
and         (8)  F  =  Ax*  +  Bx0y0  +  Cy*  +  &x0  +  Ey0  +  F. 

It  appears  then  that  we  can  choose  x0  and  y0  so  that 
any  two  of  the  last  three  terms  shall  vanish.  Let  them 
be  chosen  so  that  D'  and  E'  shall  be  zero,  or  so  as  to 
satisfy  the  two  equations, 

(9)  2AxQ  +  ByQ  +  V^0, 

and  (10)  Bx0  +  2  Oy0  +  E  =  0. 


Ch.  XIII,  §00]     EQUATION   OF   THE    SECOND    DEGREE  169 

,ri  2  CD -BE,        ,  2AE-BD 

The  general  equation  is  reduced  by  this  transformation 

(11)  Ax2  +  Bxy  +  6y  +'J"  =  0, 

in  which  the  value  of  F'  is  found  by  substituting  x0  and 
yQ  for  a;  and  y  in  equation  (1). 

But,  if  J92  —  4^1(7  =  0,  no  values  can  be  found  for  xQ 
and  y0,  and  this  transformation  is  therefore  impossible. 
Let  this  case  be  set  aside  for  the  present,  and  only  those 
cases  be  considered  where  B2  —  4  AC  ^=  0,  and  where  this 
transformation  is  therefore  possible. 

We  have  reduced  the  equation  to  a  form  which  shows 
that  the  curve  is  symmetrical  with  respect  to  the  origin, 
for  any  line,  y  =  Ix,  through  the  origin  meets  it  at  two 
points  equally  distant  from  the  origin.  But  the  term  in 
xy  must  be  removed  before  it  is  symmetrical  with  respect 
to  the  axes. 

90.    Removal   of    the   term   in   jr/.  —  Let   the   axes   be 

revolved  through  any  angle  6  by  substituting 

x  =  x1  cos  0  —  y'  sin  #, 
y  =  xr  sin  6  +  y'  cos  0. 
The  equation  now  becomes 

(12)  A'x2  +  B'xy  +  C'y2  +  F'  =  0, 
where     (13)  A'  =  A  cos2  0  +  B  sin  0  cos  0  +  C  sin2  0, 

(14)  B'  =  (C-A)am20+Bco820, 

(15)  C'  =  4sin20-£sin0cos0-f-(7cos20, 


170  ANALYTIC   GEOMETRY  [Ch.  XIII,  §  91 

Evidently  0  can  be  so  chosen  as  to  make  any  one  of 
the  three  coefficients  zero.  But  it  is  the  term  in  xy 
which  is  not  wanted.     We  shall  then  choose  6  so  that 

£'=0. 

B 


Hence  (16)  tan  2  6  = 


A-0 


There  will  always  be  two  values  of  0,  one  acute  and 
the  other  obtuse,  which  will  satisfy  this  equation.  But, 
for  the  sake  of  uniformity,  we  shall  always  choose  the 
acute  value. 

The  equation  will  be  reduced  by  this  transformation  to 

(17)  A'x2+C>y2  +  F  =  0. 

91.   Determination  of  the  coefficients  A',  C,  and  F' .  — We 

have  shown  how  to  determine  F' ;    and  since   tan  2  6  is 
known,  the  values  of  A'  and  C  may  be  found,  and  the 
result  fully  determined.     But  much  of  the  labor  involved 
may,  in  practice,  be  avoided  by  the  following  method : 
Adding  equations  (13)  and  (15),  we  have 

(18)  A'  +  C  =  A  +  a 

Subtracting  the  same  equations,  we  have 

(19)  A'-C'=(A-  67) cos  2  (9-f-^ sin  2 (9. 

Squaring  (19)  and  (14),  and  adding,  we  have 

(20)  (A!  -  Cr)2  +  B12  =  (A-  O)2  +  B2. 

Squaring  (18)  and  subtracting  from  (20),  we  have 

(21)  B'2-4ArC'  =  B2-4,AO. 

These  results  hold  good  for  all  transformations  from 
one  system  of  rectangular  axes  to  another. 


Ch.  XIII,  §  91]     EQUATION   OF   THE   SECOND   DEGREE  171 

If  the  general  equation  has  been  reduced  to  equation 
(IT),  B'  =  0,  and  (21)  reduces  to 

(22)  4A'Cf=4tAC-B*. 

From  the  two  equations  (18)  and  (22),  A'  and  C  can 
be  found.  But  there  will  be  two  values  of  each,  corre- 
sponding to  the  two  possible  values  of  0,  and  it  will  be 
necessary  to  be  able  to  choose  the  proper  values. 

We  have  let  (C-  A)  sin  2  6  +  £cos  2  6  =  0. 

Multiplying  this  equation  by  (^4.  —  (7)  and  equation 
(19)  by  B  and  subtracting,  we  have 

(23)  (B2  +  (A-  C)2)  sin  2  0  =  B(A'  -  C). 

If  now  the  acute  value  of  6  be  chosen,  the  first  member 
will  always  be  positive,  and  the  factors  of  the  second 
member,  B  and  A'  —  C,  must  have  the  same  sign.  It  will 
be  easy  then  to  choose  the  proper  values  for  A'  and  C '. 

The  determination  of  F'  may  also  be  considerably  sim- 
plified. Multiply  equation  (9)  by  x0  and  (10)  by  y0  and 
add.     The  sum  is 

2  Ax*  +  2 Bx0y0  +  2  Cy*  +  Dx0  +  Eyn  =  0. 

Combining  this  with  (8),  we  have 

■*"  =  f*o  +  fs'o  +  -*r- 

Substituting  the  values  of  x0  and  y0, 

CD2  +  AE2  -  BDE+  B*F-  4  A  OF  =       -A 

B2-±AC  tf-lAQ 


(24)  F'  = 


172  ANALYTIC    GEOMETRY  [Ch.  XIII,  §  92 

92.    Nature  of  the  locus.  —  The  general  equation  has 

now  been  reduced  by   transformation   of   coordinates  to 

the  form 

(IT)  A'x2  +  C'y2  +  F'  =  0. 

Neither  of  the  coefficients  A!  or  C  can  be  zero,  for 
they  must  satisfy  equation  (22),  and  we  are  only  con- 
sidering the  case  where  B2  —  -±AC^0.  If  F'  =£  0,  (17) 
can  be  written  in  the  form 

-F'  "*"  -F'~ 

~aJ~     a 

The  nature  of  the  curve  evidently  depends  on  the  rela- 
tive signs  of  A',  C' ',  and  F'.  If  A'  and  C  have  the  same 
sign  and  F'  has  the  opposite  sign,  equation  (17)  will 
represent  a  real  ellipse;    for   it    can    be   written    in   the 

form 

^  +  £=1 
a2      b2 

If  A',  C1 ',  and  F'  all  have  the  same  sign,  equation  (17) 
can  be  written  in  the  form 

^  +  ^  =  -  1 
a2^b2 

This  equation  has  no  real  locus,  but  is  said  to  represent 
an  imaginary  ellipse. 

Again,  if  A1  and  C  have  opposite  signs,  the  equations 
will  take  one  of  the  two  forms 

^_^2-1     or     ^-t-      \ 
a2      b2~    '  a2      62~        ' 


Ch.  XIII,  §92]     EQUATION    ()K  THE   SECOND   DEGREE         173 

according1  as  F'  has  the  same  sign  as  C  or  as  A'.  These 
are  both  real  hyperbolas. 

From  equation  (22),  4  A'  C  =  4  A  C  —  B2,  we  see  that 
A'  and  C  have  the  same  or  opposite  signs  according  as 
B2  —  4:  AC  is  negative  or  positive.  If  then  F'  is  not  zero, 
or  what  is  the  same  tiling,  if  the  discriminant  does  not 
vanish,  and  if  B2  —  \  AC  ^  0,  the  general  equation  has 
been  shown  to  represent 

an  ellipse,  real  or  imaginary,  when  B2  —  4  AC <  0, 

an  hyperbola,  always  real,       when  B2  —  4  AC  >  0. 

If  F'  =  0,  equation  (17)  reduces  to 

(25)  A'x2  +  C'y2  =  0. 

When  A'  and  C  have  the  same  sign,  the  equation  may 
be  looked  upon  as  representing  a  pair  of  imaginary  lines, 
since  the  equation  can  be  separated  into  a  pair  of  linear 
factors  with  imaginary  coefficients.  These  lines  have  a 
real  point  of  intersection,  the  origin.  Or  it  may  be  looked 
upon  as  the  equation  of  an  ellipse  in  which  the  axes  have 
become  zero.  From  this  point  of  view,  it  is  spoken  of  as 
representing  a  null  ellipse. 

•  When  A!  and  C  have  opposite  signs,  the  equation  can 
always  be  separated  into  two  real  factors,  representing 
two  real  and  intersecting  lines. 

These  results  will  be  seen  to  agree  with  those  obtained 
in  Art.  88. 

PROBLEMS 

1.  Determine  the  character  of  the  locus  of  the  following 
equation,  reduce  it  to  its  simplest  form,  and  plot : 

5x2  +  2xy  +  5y2-12x-l2y  =  0. 


174  ANALYTIC   GEOMETRY  [Ch.  XIII,  §  92 

Substituting  these  values  for  the  coefficients  in  (4),  we 
obtain  A  =  -1152.  Also  B2-4AC  =  -96.  The  locus  is 
therefore  an  ellipse,  real  or  imaginary. 

The  simplest  method  for  determining  the  coordinates  of  the 
centre  is  to  write  the  equations  (9)  and  (10)  and  solve  for 
x0  and  y0. 

The  first  of  these  may  be  obtained  by  multiplying  the  coeffi- 
cient of  every  term  which  contains  x  by  the  exponent  of  x, 
decreasing  that  exponent  by  unity,  and  leaving  out  all  terms 
which  do  not  contain-  x.  The  second  may  be  formed  in  a 
similar  way,  using  y.     In  this  case  they  are 

10  x0  +  2y0  -  12  =  0,  and  2  x0  +  10yQ  -  12  =  0. 

From  these  the  coordinates  of  the  centre  are  found  to  be  (1, 1). 
From  equation  (24),  F'  =  —  12.     The  equation,  referred  to 
the  point  (1, 1)  as  origin,  is  then 

Sx2  +  2xy  +  5y2  -12  =  0. 

Next  revolve  the  axes  through  an  angle  6,  such  that 

tan  2,  =  -^  =  ^. 

We  have  decided  to  use  the  acute  value  of  6,  which  is  here  -• 

4 

To  determine  A!  and  C,  we  use  the  equations  (18)  and  (22) 

or  Ar+C'  =  A  +  C  =  10, 

and  4  A'C  =  4  AC  -  B2  =  96. 

Solving,  we  have  A'  =  6  or  4,  and  C  =  4  or  G.  But  since 
we  chose  the  acute  value  of  6,  we  must  choose  A'  and  C  so  that 
A'  —  C  has  the  same  sign  as  B.  This  is  positive.  Hence  the 
final  form  of  the  equation  is 

6x2+4y2  =  12. 


Ch.  XIII,  §  03]     EQUATION   OF   THE   SECOND   DEGREE         175 


But   this  is  the   equation   of   the    curve   referred    to  axes 
with    origin   at   the   point 
(1, 1),  and  making  an  angle 

of  -  with  the  original  axes. 
4 

Constructing  these  axes 
and  plotting  the  equation 

Gar2 +  4/ =  12 
with  respect  to   them,  we 
have  the  locus  of  the  origi- 
nal equation,  referred  to  the 
original  axes. 

2.    Determine  the  char-  FlG-  90- 

acter  of  the  loci  of  the  following  equations,  reduce  them  to 
their  simplest  forms,  and  plot : 

(a)  2a2 +  2f-  4^-47/4-1  =  0, 

(b)  x*  +  tf  +  2x+2  =  0, 

(c)  ±xy- 2x  +  2=0, 

(d)  y2-5xy  +  6x2-Ux  +  5y  +  4:  =  0. 

Case  II.     B2-AAC  =  0. 

93.  Removal  of  the  term  in  xy.  —  We  have  seen  that,  if 
B2  —  4  A  0  =  0,  it  is  not  possible  to  transform  to  a  new 
origin  such  that  the  terms  in  x  and  y  shall  disappear.  In 
this  case  we  shall  first  revolve  the  axes  through  an  angle  0. 

Proceeding  as  in  Art.  90,  we  obtain  the  equation 

A'x2  4-  B'xy  4-  O'f  4-  D'x  +  E'y  +  F=0, 
where     (13)  Af  =  A  cos2  6  4-  B  sin  6  cos  6  4-  tfsin2  0, 

(14)  B'  =  (C  -  A)  sin20  +  Bcos20, 

(15)  C  =  Asin20  -Bsin0cos0  +  Ccos20, 

(26)  B'  =  Bcos0  +  Esin0, 

(27)  E'  =  -  2)  sin  0  +  E  cos  0, 


176  ANALYTIC   GEOMETRY  [Ch.  XIII,  §  93 

The  values  of  A',  B\  and  C  are  the.  same  as  those  used 
in  Art.  91.  The  results  there  obtained  will  therefore 
apply  here.     These  were, 

(18)  A'  +  C  =  A+  C, 

and  (21)  B't-lA'C'^B't-'iAa 

Let  6  be  so  chosen  that  B'  =  0,  or  tan  20  =  — -•    Then 

A  —  C 

since  B2  —  \AQ  =  0,  it  appears  from  (21)  that  either  A! 
or  C  must  reduce  to  zero  at  the  same  time.  It  can  easily 
be  shown  that  one  of  the  two  values  of  6  will  give  A'  =  0, 
and  the  other,  C  =  0.  Let  that  value  be  chosen  which 
will  make 

A'  =  A  cos2  6  +  B  sin  6  cos  0  +  C  sin2  0  =  0. 

Solving,  we  have     A  +  B  tan  6  +  C  tan2  (9  =  0, 

The  general  equation  will  be  reduced  by  this  transforma- 
tion to 

(29)    O'f  +  D'x  +  B'y  +  F=Q, 

where  (30)    C  =  A  +  C, 

(3n^=   BD-2AH% 

±V^2  +  4A2 

and  (32)^=    ^B  +  2AD 


±  V£2  +  4  A2 


It  appears  then  that  C  cannot  vanish,  since  A  and 
Q  have  the  same  sign  ;  that  D'  or  E'  may  vanish,  but 
since  BD  —  2  A E  is  the  value  of  the  discriminant  when 
B2  —  4  J.  (7=  0,  I)'  cannot  be  zero  unless  A  =  0. 


Ch.  XIII,  §  95]     EQUATION  OF   THE   SECOND   DEGREE  177 

94.  Removal  of  the  term  in  /.  —  Transform  equation 
(29)  to  a  new  origin  (:r0,  ?/0).     It  becomes 

(33)    O'f  +  D'x  +  E"y  +  F'  =  0, 

where  (34)  E"  =  2  O'y0  +  E\ 

and  (35)  F>  =  C'y*  +  D'xQ  +  E'y0  +  F. 

We  can  then,  in  general,  choose  such  values  for  x0  and  yQ 
that  E"  =  F'  =  0.     Solving  the  two  equations 

and  C'y*  +  D%  +  E'i,0  +  F=0, 

,                               E'         ,          J?'2  -  4  67'^ 
we  have  y0  =  -  — ,  and  x0  =       ^QJJ)] 

If  D'  =£  0,  these  values  are  always  finite,  and  the  trans- 
formation is  possible.     The  equation  will  be  reduced  by  it  to 

(36)  C'f  +  D'x  =  0. 

If  D'  =  0,  no  value  can  be  obtained  for  x0  which  will 
make  F'  =  0.  But  if  we  transform  to  the  point  (0,  ?/0), 
the  equation  will  be  reduced  to 

(37)  C'y2  +  F'  =  0. 

95.  Nature  of  the  locus.  —  When  B2-4AO=0,  the 
general  equation  has  been  reduced  by  transformation  of 
coordinates  to  one  of  the  two  forms 


(36)   Cy  +  3'x=0,OTtf=--%;X% 


D[ 

C 


(37)   C'f  +  Ff  =  0,oTf  =  --^ 


178  ANALYTIC   GEOMETRY  [Ch.  XIII,  §  90 

The  first  of  these  equations  always  represents  a  real 
parabola.  The  second  is  obtained  only  when  A  =  0,  and 
represents,  as  we  should  expect,  a  pair  of  lines.  In  this 
case  the  lines  are  evidently  parallel  and 

real  and  distinct,        if  C  and  F!  have  opposite  signs, 

real  and  coincident,  if  F'  =  0, 

imaginary,  if  (7' and  F'  have  the  same  sign. 

It  has  been  shown  that 

(30)    C'  =  A  +  C, 

and  (31)  D>=-ZV-2AE  , 

and  when  A  =  0,  it  can  be  shown  that 

4  AF  -  D2 


(38)  F'  = 


±A 


From  these  values  the  reduced  form  of  the  equation  can 
be  determined.  But  in  any  numerical  problem  the  method 
of  the  following  section  will  be  found  to  be  simpler. 

PROBLEM 

1.  Show  that  the  above  conditions  which  determine  the 
nature  of  the  parallel  lines  are  the  same  as  those  given  at  the 
end  of  Art.  88. 

96.  Second  method  of  reducing  the  general  equation  to  a 
simple  form,  when  B2  -  4  AC  =  0.  —  When  B1  -  4  AO=  0, 
the  terms  of  the  second  degree  in  the  general  equation 


Ch.  XIII,  §9C]     EQUATION  OF  THE   SECOND   DEGREE  179 

form  a  perfect  square,  and  the  equation  can  be  written  in 

the  form 

(ax  +  cy)2  +  Dx  +  Ey  +  F  =  0, 

where  a  =  V^L  and  e  =  V(7. 

Introduce  arbitrarily  the  quantity  k  inside  the  paren- 
thesis, and  subtract  from  the  rest  of  the  equation  whatever 
has  been  added  by  this  introduction.     It  becomes 

(ax  +  cy  +  7c)2  +  (D  -  2 ak)x  +  (E-2  ck)y  +  F-k2  =  0. 

Now  choose  such  a  value  for  k  that  the  two  lines  repre- 
sented by  the  equations 

ax  -f-  cy  +  k  =  0 

and  (D-  2  a¥)x  +  (E-  2  ck)y  +  F-k2  =  0 

shall  be   perpendicular  to   each  other.     Let  I  be  such  a 
value  of  k.     The  equation  will  then  take  the  form 

(ax  +  cy  +  02  =  D'x  +  E'y  +  Ff. 

Divide  both  members  of  this  equation  by  a2  +  c2, 
and  both  divide  and  multiply  the  second  member  by 
VD/2  +  E'2,  and  write  the  result  in  the  form 


fax  +  cy  +  W=  Vi>'2  +  E'2  (D'x  +  E'y  +  F'\ 
\  ^a^+72  J  <#  +  <?      \    VI)'2  +  E'2    J 

But    x       y  —  is  the  distance  of  the  point  (x,  y)  from 
Vfl2  +  c2 
the  line  ax  +  cy  +  I  =  0.     Represent  this  by  y1 . 


180  ANALYTIC  GEOMETRY  [Ch.  XIII,  §96 

Again,   — x  ^  — —   is   the    distance    of    the   point 

v D'2 + m 

(x,  ?/)  from  the  line  D'x  -\-  E'y  +  F'  =  0.     Represent  this 
by#'.     Then  the  equation  reduces  to 


n_(y/D'*  +  E'*\_l 


where  x'  and  yf  represent  the  perpendicular  distances  of 
any  point  on  the  locus  from  the  two  perpendicular  lines 

D'x  +  E'y  +  F>  =  0 

and  ax  +  cy  -f  I  =  0. 

It  is  therefore  the  equation  of  the  curve  referred  to 
these  lines  as  Y  and  Jf-axes  respectively.  The  positive 
direction  of  the  X-axis  can  be  fixed  by  finding  the  inter- 
cepts of  the  curve  on  the  original  axes,  and  determining 
by  inspection  which  way  the  parabola  is  turned. 

PROBLEMS 

1.    Plot  the  locus  of  the  equation 

tf  _  2  xy  +  y2  -  8  x  +  16  =  0. 

Following  the  method  described  above,  write  the  equation  in 
the  form 

(x  —  y  +  k)2  -(8  -f-  2  k)x  +  2  ky  + 16  -  k2  =  0. 

If  the  two  lines  represented  by  the  equations 

x  —  y  +  k  =  0 

and  _(8  +  2  ft) a;  -f  2  %  +  16  -  A:2  =  0 


Ch.  XIII,  §96]     EQUATION   OF  THE   SECOND   DEGREE 


181 


are  perpendicular  to  each  other,  k  =  —  2.     Substituting  this 
value  in  the  equation  and  transposing,  it  becomes 

(x-y-2y  =  4:(x  +  y-3). 

Dividing  both  members  by  a2  +  c2,  and  both  dividing  and 
multiplying  the  second  member  by  V ' Dn  -f  E'2,  it  becomes 


(^)'=2v2(^) 


V2{X  +  y-°\  or  y'2=2V2x', 


where  y'  is  the  perpendicular  distance  of  any  point  (#,  y)  of 

the  locus  from  the  line   x  —  y  —  2  =  0,  and  where  x'  is  the 

distance  from  x-\-y—S=0. 

It  is  therefore  the  equation 

of    the    locus   referred    to 

these  lines   as   X  and    Y- 

axes. 

Construct  the  two  lines. 
From  the  original  equation 
we  see  that  the  curve 
touches  the  X-axis  at 
the  point  (4,  0),  and  does 
not     cut    the     F-axis.     It 

is  then  easily  seen  which  FlG-  91- 

is  the  positive  direction  of  the  axis  O'X',  and  the  curve  can  be 
plotted  as  in  Fig.  91. 

2.    Plot  by  this  method  the  locus  of  the  following  equations: 
(a)  x2  -  2  xy  +  y2  -  6  x  -  6  y  +  9  =  0, 
(6)  x2  +  6  xy  -f-  9  y2  +  x  -  6  y  -  9  =  0, 

(c)  2x2  +  $y2  +  8xy  +  x  +  y  +  3  =  0, 

(d)  f  -  2  x  -  8  y  +  10  =  0, 

(e)  4:X2  +  4xy  +  y2  +  6  =  0. 


183 


ANALYTIC   GEOMETRY 


[Ch.  XIII,  §  97 


when  A  =  0,  • 


97.  Summary.  —  It  has  been  shown  in  this  chapter  that 
the  general  equation  of  the  second  degree  represents, 

'  and  when  B2  —  4  AC  <  0,  an  ellipse  (real  or 

imaginary), 
when  A  *  0,  j  ftnd  when  &  _  4  A  Q  =  ^  ft  parabola? 

and  when  B1  —  4  A  C  >  0,  an  hyperbola ; 

and  when  B2  —  4  AC <  0,  a  null  ellipse  (two 
imaginary  lines), 

and  when  B2  —  4  A  C  =  0,  two  parallel  lines 
(real,  coincident,  or  imaginary), 

and  when  B2  —  4  A  C  >  0,  two  real  intersect- 
ing lines. 

All  of  these  forms  may  be  obtained  as  plane  sections  of 
a  right  circular  cone,  and  are  all  included  under  the  term 
"conies."  An  equation  of  the  second  degree  must  therefore 
represent  some  conic  either  in  its  regular  or  degenerate  form. 

PROBLEMS 

Determine  the  nature  of  the  locus  of  each  of  the  following 
equations : 

1.  3x?-2xy  +  y2  +  2x  +  2y  +  5  =  0. 

2.  x2  +  xy  +  y2  +  2  x  +  3y  -3  =  0. 

3.  2X2-  5xy-  3/4-9  x-  13  y  +  10  =  0. 

4.  4z2  +  2xy-\f  +  6x  +  2y  +  3  =  0. 

5.  9x2-  12xy  +  ±tf-  24  a;  +  16  y-  9  =  0. 

6.  9arJ-6a-?/  +  ?/2  +  4a;  +  3?/  +  16  =  0. 

7.  25  x2  +  40  xy  +  16  y2  +  70  x  +  56  y  +  49  =  0. 

8.  13  a;2  +  14  ajy  +  5  y2  +  14  x  +  10  y  +  5  =  0. 

9.  4  x2  +  9  t/2  -  8  a?  +  54  y  +  85  =  0. 

10.    3a:2  +  10a^  +  7?/2  +  4x-f-2?/  +  l  =  0. 


Cm.  XIII,  §99]     EQUATION  OF  THE   SECOND  DEGREE  183 

98.  General  equation  in  oblique  coordinates.  —  If  the 
general  equation  is  referred  to  axes  which  are  oblique, 
we  can  first  transform  to  rectangular  axes  with  the  same 
origin.     The  resulting  equation  will  be  in  the  form 

A'x2  +  B'xy  +  Cif  +  D'x  +  E'y  +  F'  =  0. 

This  can  then  be  treated  by  the  methods  of  this  chapter. 
It  must,  therefore,  represent  a  conic. 

99.  Conic  through  five  points.  —  The  general  equation 
of  the  second  degree  contains  six  constants,  but  only  five 
of  these  are  independent,  since  any  one  we  please  may  be 
reduced  to  unity  by  division.  Five  conditions  are  therefore 
sufficient  to  determine  the  conic.  For  example,  it  can  be 
made  to  pass  through  five  points,  and  in  general  no  more 
than  five.  For,  if  the  coordinates  of  the  five  points  are 
substituted  in  turn  in  the  general  equation,  there  will  be 
five  equations  from  which,  in  general,  we  can  determine 
five  coefficients  in  terms  of  the  sixth,  which  will  divide 
out  after  substitution.  If  a  sixth  point  were  given,  there 
would  be  six  simultaneous  equations  in  five  variables, 
which  is  not  possible  unless  some  of  the  equations  are  not 
independent.  This  will  only  happen  when  the  sixth  point 
lies  on  the  conic  through  the  other  five. 

If  three  of  the  points  lie  on  a  line,  the  conic  evidently 
breaks  up  into  this  line  and  another  line  through  the  other 
two.  If  four  points  lie  on  a  line,  the  solution  is  indeter- 
minate ;  for  this  line  and  any  other  through  the  fifth 
point  will  be  a  conic  through  the  five  given  points. 

Other  conditions  may  be  given,  as  in  the  case  of  the 
circle,  where  A  =  C  and  B=  0.     These  two   conditions 


184  ANALYTIC   GEOMETRY  [Ch.  XIII,  §99 

restrict  the  number  of  points  through  which  a  circle  can 
be  passed  to  three.  Similarly,  a  parabola  can  be  passed 
through  only  four  points,  since  the  condition  B1  —  4  AC=Q 
must  be  satisfied.  But  here,  since  the  condition  is  a  quad- 
ratic, there  may  be  two  parabolas  which  pass  through  the 
four  points ;  or  imaginary  solutions  may  be  obtained,  and 
four  points  may  therefore  be  chosen  through  which  no  real 
parabola  can  be  drawn. 

PROBLEMS 

1.  Find  the  equation  of  a  conic  through  the  points 

(a)  (2,  3),  (0,  -  3),  (2,  0),  (5,  5),  (-  5,  -  5). 
(6)   (5,  3),  (4,  4),  (2,  6),  (7,  1),  (0,  0). 
(o)    (2,  4),  (4,  3),  (6,  2),  (0,  - 1),  (1,  0). 

2.  Find  the  equation  of  a  parabola  through  the  points 

(a)  (0,0),  (8,8),  (4,2),  (-4,2). 

(b)  (0,  0),  (1,  0),  (-  1,  1),  (-  1,  -  1). 

(c)  (4,3),  (0,-4),  (6,1),  (-6,2). 

(d)  (12,-6),  (3,0),  (0,2),  (-3,4). 

3.  Determine  the  nature  of  the  conies  obtained  in  problems 
1  and  2. 


CHAPTER   XIV 

PROBLEMS    IN   LOCI 


1.    Find  the  locus  of  the  vertex  of  a  right  angle  whose 

1. 


or        ij' 
sides  are  tangent  to  the  ellipse  —  +  *- 


<(- 


The  equations  of  any  two  perpendicular  tangents  P' K 
and  P'L  may  be  written  in  the  form 


y 


Ip  +  'y/lfcP  +  lPi 


and 


t/  =  ljc  +  ^l2*a2  +  P, 


where  l1l2  =  —  1.     If  JPf  is  their  point  of  intersection,  its 
coordinates  (V,  y')  must 
satisfy  both  equations. 

Substituting  these  co- 
ordinates, and  replacing 
l2  by ,  we  have 


y  =  ^+VZ12a2-h^2, 


Fig.  92. 


two  equations  in  x\  y\ 
and  the  variable  param- 
eter lv  By  eliminating  lv  we  shall  obtain  a  single  equa- 
tion in  xr  and  yf.  Clearing  of  fractions,  transposing,  and 
squaring, 

185 


186  ANALYTIC  GEOMETRY  [Ch.  XIV 

y'2  -  2  lxx'y'  +  l2x'2  =  l2a2  +  b2 
l2y'2  +  2  Z^y  4-      x'2  =      a2  +  ^262 

Adding,  (l  +  Z12)y2+(l  +  ?12)^2  =  (l  +  ?12)a2  +  (l  +  Z12)62. 
Dividing  by  (1  +  I2),    y'2  +  a:'2  =  a2  +  62, 
or  x2  -+-  ?/2  =  a2  +  62. 

The  locus  is  the  director  circle,  a  circle  having  the  same 
centre  and  Va2  +  b2  as  radius. 

2.  Find  the  locus  of  the  intersection  of  perpendicular 
tangents  to  a  parabola. 

3.  Find  the  locus  of  the  intersection  of  tangents  to  the 
ellipse  if  the  product  of  their  slopes  is  constant. 

As  in  problem  1,  the  equations  connecting  x\  y\  lv  and 
Z2  are 


(i)   /  =  z1^+VZ12^2  +  P, 


(2)  y'  =  l2x'+Vl22a2  +  b2, 

(3)  1,1,  =  k. 

But  the  method  of  elimination  used  in  that  problem 
will  not  apply  here.     Transpose  and  square  (1)  and  (2), 

y'2  -  2  l^'y*  +  l2x'2  =  l2a2  +  b2, 

y'2  -  2  l2x'yf  +  ?2V2  =  J2a2  +  b2. 

Write  these  as  affected  quadratic  equations  in  lx  and  Z2, 

(4)  (a2  -  xf2)  I2  +  2  x'y\  +  b2-  y'2  =  0, 

(5)  (a2  -  x'2)  I2  +  2  x'y'l2  +  b2-y'2  =  Q 
If  now  we  write  the  equation 

(6)  (a2  -  x'2)  z2  +  2  x'y'z  +  b2-y'2=0 


Ch.  XIV]  PROBLEMS   IN   LOCI  187 

(an  affected  quadratic  in  2),  it  appears  from  (4)  and  (5) 
that  lx  and  l2  are  the  two  roots  of  (6),  and  hence  that 

b2  —  y'2  b2  —  y'2 

lih  =  n2      U'      But    llh  =  k-      Hence    ^-^T2  =  k    is    the 

equation    of    the    desired    locus.     Dropping   primes   and 
reducing,  we  have  kx2  —  y2  =  ka2  —  b2. 

If  k  =  —  1,  it  becomes  x2  -h  y2  =  a2  +  b2,  as  in  problem  1. 

4.  Find  the  locus  of  the  intersection  of  tangents  to  the 
parabola  if  the  product  of  their  slopes  is  constant. 

5.  Find  the  locus  of  the  feet  of  perpendiculars  from 
a  focus  on  tangents  in  the  (a)  ellipse,  (5)  hyperbola, 
(c)  parabola. 

6.  Find  the  locus  of  the  intersection  of  tangents  at  the 
ends  of  conjugate  diameters  of  an  ellipse. 

Note.  —  Solve  this  as  a  special  case  of  problem  3. 

7.  Find  the  locus  of  the  intersection  of  tangents  at 
the  ends  of  conjugate  diameters  of  an  hyperbola. 

8.  Radii  vectores  are  drawn  at  right  angles  from  the 
centre  of  an  ellipse.  Find  the  locus  of  the  intersection  of 
tangents  at  their  extremities. 

9.  Find  the  locus  of  the  middle  point  of  chords  joining 
the  ends  of  conjugate  diameters  of  an  ellipse. 

Let  (V,  y)  be  the  middle  point  of  any  such  chord.     If 

(rr^j)  are  the  coordinates  of  Pv  (  ~~,      ?  -— 1)  will  be  the 
coordinates  of  P2. 

ay.  bx* 

x*~b  y^^r 

Then         x'  = ^ »  and  V*  = 2 ' 

or  (1)  2  bx'  =  bxx  —  ayv 

and  (2)   2  ay'  =  ayx  +  bxv 


188  ANALYTIC   GEOMETRY  [Ch.  XIV 

Since  (xxy^)  lies  on  the  ellipse, 

(3)  b2x2+a2y2  =  a2b2. 

From   these   three   equations  we    can   obtain  a  single 

equation  in  x'  and  y'  by 
eliminating  x1  and  yv 
From  (1)  and  (2), 

„.    _  hx'  +  ay'_ 


*-* 


b 

'  -bx' 
a 


Fig.  93. 


Substituting  these  val- 
ues in  (3),  it  reduces  to 

a?  ,y^_  1 
'a2      b2~2 


10.  Find  the  locus  of  the  vertex  of  a  triangle  whose 
base  is  a  line  joining  the  foci  and  whose  sides  are  parallel 
to  two  conjugate  diameters. 

11.  Find  the  locus  of  the  middle  point  of  chords 
drawn  through  a  fixed  point  in  the  (a)  ellipse,  (£>) 
parabola. 

12.  Tangents  are  drawn  to  the  parabola  y2=2mx. 
Find  the  locus  of  their  pole  with  respect  to  the  circle 
x2  +  y2  =  r2. 

13.  The  two  circles  x2  +  y2  =  a2  and  x2  +  y2  —  ax  =  0 
are  tangent  internally.  Find  the  locus  of  the  centres 
of  circles  which  are  tangent  to  both  the  given  circles. 


Ch.  XIV] 


PROBLEMS   IN   LOCI 


189 


Let  the  two  circles  be  drawn,  and  let  (V,  y')  be  the 
centre  of  any  circle  which  is  tangent  to  both  circles. 
Then  the  lines  O'P'  must 
pass  through  B,  the  point 
of  contact  of  the  two 
circles,  and  OP'  must 
pass  through  0.    Hence, 

P' 0=00 -OP' 
=  r-  OP', 
and  P'B  =  P'0'-BO' 
=  P'0'  --> 


But  P'  C  and  P'B  are 

radii  of  the  same  circle. 

Hence,  r-OP'  =P'0' 


Fig.  94. 


Or 


f-v. 


x'2  4-  y'2 


4 


+  y 


'2 


Squaring  and  reducing,  the  equation  of  the  locus  re- 
duces to  8  x2  +  9  y2  —  4  rx  —  4  r2  =  0.  What  curve  is  this 
and  how  is  it  situated? 

14.  Find  the  locus  of  the  centres  of  all  circles  which 
pass  through  the  point  (0,  3)  and  are  tangent  internally 
to  x2  +  y2  =  25. 

15.  Find  the  locus  of  the  centres  of  circles  which  are 
tangent  to  a  given  circle  and  pass  through  a  fixed  point 
outside  of  that  circle. 

16.  Lines  are  drawn  from  the  point  (1,  1)  to  the 
hyperbola  x2  —  y2  =  1.  Find  the  locus  of  the  points  which 
divide  these  lines  in  the  ratio  of  2  to  1. 


190 


ANALYTIC   GEOMETRY 


[Ch.  XIV 


17.  Lines  are  drawn  from  the  centre  0  of  the  circle 
x2  +  y2  =  r2,  cutting  the  circle  in  A  and  the  line,  x  =  a, 
in  B.  Find  the  locus  of  P,  if  0,  A,  B,  and  P  form  a 
harmonic  range.  Show  that  the  result  will  represent  an 
ellipse,  hyperbola,  or  parabola,  according  as  4  r2  <  a2, 
4  r2  >  a2,  4  r2  =  a2. 

18.  Find  the  locus  of  the  vertex  of  a  triangle  if  the 
length  of  the  base  is  c,  and  the  product  of  the  tangents 
of  the  base  angles  is  k. 

Let  Pr  be  any  position  of  the  vertex  of  the  triangle, 
and  00  the  base.     We  know  that 

tan  OOP'  •  tanP'C0  =  &. 


But  tan  OOP  =  ^, 

x' 

and      tan  P' 670  = -^— -• 

c  —  x 

Hence  the  condition  which 
must  be  satisfied  is 

J2 


Fig.  95. 


tf 


=  Jc. 


x\c  -  xf) 

Dropping  primes  and  reducing,  we  have 

Jcx2  +  y2  —  hex  =  0. 

This  will  be  an  ellipse  or  hyperbola,  according  as  h  is 
positive  or  negative. 


In  either  case  the  coordinates  of 


the  centre  will  be 


&  »)• 


and  the  semi-axes  will  be  -  and 


19.    Find  the  locus  of  the  vertex  of  a  triangle  if  the 
length  of  the  base  is  c,  and  the  product  of  the  tangents 


Ch.  XIV]  PROBLEMS   IN   LOCI  191 

of  the  half  base  angles  is  k.     Show  that  the  locus  is  an 
ellipse  with  the  extremities  of  the  base  as  foci. 

Note.  — Express  the  tangents  of  half  the  base  angles  in  terms  of  the 


ft)(*  ~  O 


three  sides  by  the  aid  of  the  formula  tan  \ 

y      s(s  —  a) 

20.  Find  the  locus  of  the  vertex  of  a  triangle  if  the 
length  of  the  base  is  c,  and  one  of  the  base  angles  is  twice 
the  other. 

21.  Find  the  locus  of  the  intersection  of  tangents  to  the 
(a)  parabola,  (5)  ellipse,  (<?)  hyperbola,  which  include  an 
angle  6. 

Show  that  if  6  =  90°,  the  results  reduce  to  those  ob- 
tained in  problems  1  and  2. 

22.  Find  the  locus  of  the  centre  of  a  circle  which 
passes  through  a  fixed  point  and  touches  a  given  line. 

23.  A  straight  line,  whose  length  is  c,  slides  between 
two  perpendicular  lines.  Find  the  locus  of  the  intersec- 
tion of  the  medians  of  the  triangles  formed. 

24.  If  a  straight  line  passes  through  a  fixed  point, 
find  the  locus  of  the  middle  point  of  that  portion  of  it 
intercepted  between  two  perpendicular  lines. 

25.  Tangents  are  drawn  to  a  circle  from  a  variable 
point  on  a  given  fixed  line.  Prove  that  the  locus  of  the 
middle  point  of  the  chord  of  contact  is  another  circle. 

26.  Find  the  locus  of  the  intersection  of  a  tangent  to 
the  circle  x2  +  y2  =  a2,  and  a  perpendicular  on  the  tangent 
from  the  point  (a,  0). 

27.  Find  the  locus  of  the  poles,  with  respect  to  the 
parabola  y2=z  2  rnx,  of  tangents  to  the  parabola  y2=  —2mx. 


192 


ANALYTIC   GEOMETRY 


[Ch.  XIV 


28.  Find  the  locus  of  the  middle  points  of  chords  of 
an  ellipse  whose  poles  lie  on  the  auxiliary  circle. 

29.  Find  the  locus  of  the  intersection  of  two  perpen- 
dicular lines  which  are  tangent  respectively  to  two  con- 
focal  ellipses. 

Let  the  equation  of   the 
two  ellipses  be 

C1)     T  +  S  =  1' 

az      b2 

Since  they  are  confocal, 
the  value  of  c  will  be  the 
F  same  in  both.     Hence 

(3)  a2~52  =  a12-612. 

The  equations  of  the  tangents  to  (1)  and  (2)  are 

y  =  lx+  VPa2  +  b2, 


y 


7  T  \  72 


+v. 


Let  (a;',  ?/')  be  their  point  of  intersection. 


Then 


y'  =  lxf  +-VlW  +  b\ 


The  elimination  of  Z  will  give  a  single  equation  in  xl 
and  y'.     Transpose  and  square. 

y'2  -  2  Ix'y'  +  ZV2  =  Z2a2  +  Z>2 
Zy24-2Zrry+    z'2=   a^  +  V^2 
Adding,  (1  +  P)y'2  +  (1  +  Z2>'2  =  ax2  +  a2Z2  +  b2  +  V?2 


Ch.  XIV]  PROBLEMS   IN   LOCI  193 

J  kit  from  (3),  af  +  h2  =  «2  +  b2. 

Substituting  and  factoring,  we  have 

(1  +  l2)ij'2  +  (1  +  P)x'2  =  a\l  +  I2)  +  b2(\  +  Z2), 
ij2  +  s'2  =  a2  +  ^2, 

or,  dropping  the  primes,  we  have  for  the  equation  of  the 
locus 

x2  +  y2  =  a2  +  b2. 

30.  Find  the  locus  of  the  intersection  of  two  perpen- 
dicular lines  which  are  tangent  respectively  to  two  con- 
focal  parabolas. 

Note. — Write  the  equation  of  the  parabola  referred  to  the  focus  as 
origin,  y1  —  2  mx  +  w2,  and  obtain  the  equation  of  the  tangent  to  it  in 

terms  of  the  slope,  y  =  Ix  +  m^  +  l^> 

31.  Find  the  locus  of  the  points  of  contact  of  tangents 
drawn  from  a  fixed  point  on  the  principal  axis  to  a  set 
of  confocal  ellipses. 

32.  Find  the  locus  of  the  middle  points  of  chords  in  a 
circle,  which  are  tangent  to  an  internal  concentric  ellipse. 

Let  (V,  y')  be  the  middle  point  of  the  chord,  and 
(xv  2/j)  and  (#2,  3/2)  its  extremities.  Then  the  equation 
of  the  chord  will  be 

The  condition  which  makes  this  line  tangent  to  the 
ellipse  W'x2  +  a2y2  =  a2b2  is 

(1)  (**i-**Jfj9*-*X#  +  52. 

\    x2  —  xx    J      \x2  —  xj 

Since  (xxy^)  and  Qr2y^)  are  on  the  circle, 


194  ANALYTIC   GEOMETRY  [Cii.  XIV 

(2)  x*  +  y*  =  r\  and  (8)  <fc' +  ft1  -  **• 

Also,  (4)  x'  =  X-i±^,     and  (5)  y>  =  y-l±Ms. 

From  these  five  equations  Ave  can  eliminate  xv  yv  xv 
and  yv  and  obtain  a  single  equation  in  x'  and  y\  which 
will  be  the  equation  of  the  locus. 

33.  Given  two  concentric  ellipses,  one  within  the  other, 
on  the  same  axes.  Find  the  locus  of  the  pole  of  tangents 
to  the  inner  with  respect  to  the  outer. 

34.  Find  the  locus  of  the  middle  points  of  a  set  of 
parallel  chords  intercepted  between  an  hyperbola  and  its 
conjugate. 

35.  Normals  are  drawn  to  an  ellipse  and  the  circum- 
scribing circle  at  corresponding  points.  Find  the  locus 
of  their  point  of  intersection. 

36.  A  perpendicular  is  drawn  from  a  focus  of  an  ellipse 
to  any  diameter.  Find  the  locus  of  its  intersection  with 
the  conjugate  diameter. 

37.  Find  the  locus  of  the  middle  point  of  all  chords  of 
an  ellipse  of  the  same  length  2  c. 

Note.  — Find  the  polar  equation  of  the  ellipse  referred  to  the  point 
(sc',  y')  as  origin.  Then  express  the  conditions  that  the  two  values  of  p 
are  each  equal  numerically  to  c,  but  opposite  in  sign.     Eliminate  0. 

38.  Find  the  locus  of  the  intersection  of  the  ordinate 
of  any  point  of  an  ellipse,  produced,  with  the  perpendicu- 
lar from  the  centre  to  the  tangent  at  that  point. 


CHAPTER   XV 


HIGHER   PLANE   CURVES 

100.  Introduction. — Any  locus  which  lies  wholly  in  a 
single  plane,  and  which  cannot  be  represented  by  an 
algebraic  equation  of  the  first  or  second  degree,  is  spoken 
of  as  a  higher  plane  curve.  Their  equations  may  be  purely 
algebraic,  or  they  may  involve  functions  other  than  alge- 
braic, when  they  are  spoken  of  as  transcendental.  There  are 
an  infinite  number  of  such  curves ;  but  only  a  few  of  those 
which  are  of  importance  in  y 

the  study  of  the  Calculus 
will  be  discussed  here. 


101.    The  parabolas.  — 

The  locus  of  any  equa- 
tion of  the  form  y  =  axn 
is  called  a  parabola  of  the  T> 
/7th  degree.  It  is  usual 
to  restrict  n  to  values 
greater  than  unity.  If 
n  =  2,  we  have  the  ordi- 
nary parabola  along  the 
F-axis.  If  n=  3,  the 
locus  is  called  the  cubi- 
cal parabola,  and  has  the 
form  shown  in  Fig.  97.     If  n 


Fig.  97. 

|,  the  locus  is  called  the 
semicubical  parabola,  and  has  the  form  shown  in  Fig.  98. 


105 


19G 


ANALYTIC   GEOMETRY 


[Ch.  XV,  §  102 


All  parabolas  are  similar  to  one  of  these  three  forms 
according  to  the  value  given  to  n. 

Let  the  student  plot  the  locus 
for  various  values  of  n.  Let  him 
also  show  that,  if  n  is  an  even 
integer,  or  a  fraction  with  an  even 
numerator  and  an  odd  denomina- 
tor, the  curve  is  similar  to  the 
ordinary  parabola ;  if  n  is  an  odd 
integer,  or  a  fraction  with  an  odd 
numerator  and  an  odd  denomina- 
tor, the  curve  is  similar  to  Fig.  97  ; 
while,  if  n  is  a  fraction  with  an 
odd  numerator  and  an  even  de- 
nominator, the  curve  is  similar 
to  Fig.   98. 

102.  The  Cassinian  oval. —The 
locus  of  a  point,  the  product  of 
whose  distances  from  two  fixed  points  is  constant,  is 
called  a  Cassinian  oval.  The  two  fixed  points  are  called 
the  foci  of  the  oval. 
To  find  its  rectangular 
equation,  let  the  X-axis 
go  through  the  two  foci, 
F  and  F,  and  let  the 
T-axis  bisect  FF .  Take 
OF  equal  to  e,  and  let 
(x,  y)  be  the  coordi- 
nates of  P,  any  point 
on  the  locus. 


Fig.  98. 


Ch.  XV,  §  102] 


HIGHER   PLANE   CURVES 


197 


Then  FP  =  V(z  -  c)2  +  ~y\  and  F'P  =  V(x  +  c)2  +  f. 
But  FP.F'P  =  m2. 

Hence       [0  -  c)2  +  f]  [O  +  c)2  +  y2]  =  m4, 


or 


(a?  +  y2  +  c2)2  -  4  c2x2=m* 


is  the  equation  of  the  Cassinian  oval. 

The  intercepts  of  the  curve  on  the  axes  are  ±  Vc2  ±  m2 
and  ±  Vm2  —  c2.  Hence  if  c<m,  the  curve  cuts  each  axis 
in  two  real  points  and  has  the  form  shown  in  Fig.  99. 
While  if  e>ra,  the  curve  cuts  the  X-axis  in  four  real 
points  but  does  not  cut  the  I^-axis.  It  must,  therefore, 
consist  of  two  distinct  ovals,  as  shown  in  Fig.   100. 

If  c  =  m,  all   the   intercepts   are   zero   and   the   curve 


Fig.  100. 


goes  through  the  origin.      In  this  case  the  equation  re- 
duces to 

x2  +  y2  =  2  c2(x>  -  y2), 

or  in  polar  coordinates, 

p2=2c2  cos  2  6. 

This  special  form  of  the   Cassinian  oval  is  called  the 
lemniscate.     It  has  already  been  discussed  on  page  67. 


198 


ANALYTIC   GEOMETRY 


[Ch.  XV,  §  103 


103.  The  cissoid.  —  The  cissoid  may  be  defined  as 
follows  :  on  any  diameter  OA  of  a  circle,  lay  off  equal  dis- 
tances CM  and  CN  on  each  side  of  the  centre,  and  at  the 
points  M  and  N  erect  MK  and  NL  perpendicular  to  the 

diameter.  Draw  OK  and  OL. 
The  locus  of  the  intersection  of 
OK  with  NL  and  OL  with  MK 
is  the  cissoid  of  Diodes. 

To  obtain  its  rectangular  equa- 
tion, let  OA  be  the  X-axis  and  0 
the  origin.  Then  031=  NA  =  x, 
and 


NL  =  VON-  NA  =  V(2  a  -  x)x. 

From  the  similarity  of  the  tri- 
angles OMP  and  ONL, 

OM:  ON::  MP :  NL, 


or  x : 


:  2  a  —  x  :  :  y  :  V(2  a  —  x): 


Fig.  101. 


Hence     y1  =  - 


za  —  x 


which  is  the  rectangular  equation  of  the  cissoid.  It  is 
evidently  symmetrical  with  respect  to  the  X-axis,  and 
has  the  line  x  =  2a  as  an  asymptote. 


104.  The  conchoid.  —  Let  A  be  a  fixed  point  at  a  dis- 
tance a  from  a  fixed  line  OX.  Draw  the  line  AP 
through  A  cutting  OX  at  B,  and  on  this  line  lay  off  a 
constant  distance  BP(=b)  both  ways  from  B.  The 
locus  of  P  is  called  the  conchoid  of  Nicomedes. 

To  find  its  rectangular  equation,  take  the  fixed  line 


Ch.  XV,  §104]  HIGHER    PLANE   CURVES 


199 


OX  as  the  X-axis,  and  OA  as  the  F"-axis.  Drop  a  per- 
pendicular from  P  on  the  X-axis  and  continue  it  to  meet 
AK,  drawn  parallel  to  the  same  axis.     Then 


But 

Hence 


or 


AP  =  V*2  +  (#  +  a)2,  and  #P  =  £. 

AP=KP 
BP     MP 

g2-K,y  +  fl)2=Q/  +  a)2 
62  y2 

Y 


The  fixed  point  ^1  is  called  the  pole,  and  the  fixed  line 

OX  the  directrix  of  the  conchoid.     If  a  <  b,  the  curve  has 

the  form  shown  in  Fig.  102.     If  a  —  5,  there  is  no  loop, 

but  the  curve  has  a  cusp  at  A.     If  a  >  5,  the  lower  branch 

of  the  curve  cuts  the  F-axis  in  a  single  point  above  A. 

Note. — Among  the  most  noted  problems  of  the  ancient  mathema- 
ticians were  the  Trisection  of  an  Angle  and  the  Duplication  of  the  Cube 
by  the  aid  of  ruler  and  compass  alone.  It  has  lately  been  shown  that 
the  solution  of  these  problems  in  this  way  is  impossible.  Both  problems 
involve  the  solution  of  a  cubic  equation,  and  both  may  be  made  to  depend 
upon  the  construction  of  two  mean  proportionals  between  two  straight 
lines.  This  has  been  accomplished  in  various  ways  by  aid  of  higher 
plane  curves,  and  it  was  for  this  purpose  that  both  the  Conchoid  and 
Cissoid  were  invented. 


200 


ANALYTIC   GEOMETRY 


[Ch.  XV,  §  10/ 


105.  The  cycloid. — The  path  described  by  a  point  on 
the  circumference  of  a  circle  which  rolls  on  a  straight 
line  is  called  a  cycloid. 

Let  C  be  the  centre  of  the  moving  circle  of  radius  a, 
and  let  P  be  the  fixed  point  on  its  circumference.  To 
find  the  rectangular  equation  of  the  curve,  let  the  X-axis 
coincide  with  the  fixed  line,  and  choose  as  the  origin  of 
coordinates  one  of  the  points  where  P  coincides  with 
that   line.     Draw   CM  perpendicular   to    OX,    and   PK 


Fig.  103. 

parallel  to  the  same  line.     Let  6  represent  the  circular 

measure   of   the   angle   MCP  through   which   the   radius 

CP  has  revolved.     Let  the  coordinates  of  P  be  (x,  y). 

Then 

x  =  OM-  PK,  and  y  =  MC  -  KC. 

But  0M=  arc  MP  =  a0, 

PK=  PC  sin  0, 

and  KC  =  PC  cos  d. 

Hence  x  —  a  (6  —  sin  0), 

y  =  a(l  —  cos  0). 

If  0  is  eliminated  from  these  two  equations  by  finding 


Ch.  XV,  §  106] 


HIGHER   PLANE    CURVES 


201 


its  value  in  the  second  equation  and  substituting  in  the 
first,  we  have 

x  =  a  vers-1  ( - )  —  V2  ay  —  y2. 


But  this  single  equation  is  not  so  convenient  to  use  as 
the  pair  of  equations  from  which  it  was  obtained.  These 
two  equations,  containing  a  third  variable,  are  equivalent 
to  the  single  equation  from  which  6  has  been  eliminated. 
The  locus  consists  of  an  infinite  number  of  branches, 
similar  to  the  one  shown  in  Fig.  103,  extending  both  to 
the  right  and  to  the  left  of  the  origin. 


106.  The  hypocycloid. — The  path  described  by  a  point  on 
the  circumference  of  a  circle  which  rolls  on  the  inside  of  a 
fixed  circle  is  called  a  hypocycloid.  Let  a  be  the  radius  of 
the  fixed  circle, 
and  b  the  radius 
of  the  rolling  cir- 
cle. Let  P  be 
the  fixed  point  on 
the  rolling  circle. 
Take  the  centre 
0  of  the  fixed 
circle  as  the  ori- 
gin of  coordinates 
and  let  the  X-axis 
pass  through  A, 
one  of  the  points 
where     P     coin-  FlG-  104- 

cides  with    the   fixed    circle.      Consider   the    rectangular 
coordinates  (x,  y)  of  any  position  of  P.     Drop   perpen- 


a  x 


202  ANALYTIC   GEOMETRY  [Ch.  XV,  §  106 

diculars  from  C  and  P  to  the  X-axis,  and  through  P  draw 
LR  parallel  to  that  axis.  The  radius  OK  of  the  fixed 
circle,  drawn  through  the  point  of  contact  K,  passes 
through  the  centre  C  of  the  rolling  circle.  Let  Z  A0K=  $, 
and  ZPCK=  6.     Then,  since  the  arcs  AK  and  PK  are 

equal,  acf)  =  bd,  or  6  =  -  </>. 
o 

Now         x  =  031=  ON-  PL, 
y  =  MP  =  NC-LC. 

But  ON  =00 cos  <f>  =  (a  -  5)  cos  <£, 
iVrC=  0(7 sin  (j>  =  (a-b)  sin  <£, 
PZ  =  PC7  cos  RPC=b  cos  [180°  -(0-  £)] 

=  —  b  cos  (6  —  $)=  —  b  cos  f  — — —  j c/>, 
and  LC=  b  sin  (0  —  <£)=  5  sin  (— -jr— -  )<£• 

Then        x  =  (a  —  5)  cos  $  +  6  cos  f  — — —  ]  <£, 
y  =  (a  —  b~)  sin  $  —  b  sin  (     ~~     )  <f>. 

As  in  the  cycloid,  these  two  equations,  containing  a 
third  variable  </>,  may  be  used  in  place  of  a  single  equa- 
tion in  x  and  y  to  represent  the  curve. 

The  most  important  special  case  is  the  four-cusped  hypo- 
cycloid,  in  which  a  =  4  b.     The  equations  here  reduce  to 

x  =  |  a  cos  <f>  +  \a  cos  3  <£, 

y  =  |  a  sin  <j>  —  |  a  sin  3  (/>. 

But         sin  3  <£  =  3  sin  </>  —  4  sin3  c/>, 

cos  3  <£  =  4  cos3  </>  —  3  cos  <j>. 


Ch.  XV,  §  107] 

Hence  x  =  a  cos3  </>, 
and  y  =  a  sin3  (f>. 

Raising  to  the  §  power 

and  adding-,  we  have 

Xs  +  y%  =  «* 
as  the  rectangular  equa- 
tion of  the  four-cuspecl 
hypocycloid.  The  form 
of  the  curve  is  shown  in 
Fig.  105. 


HIGHER   PLANE   CURVES 


203 


107.    The  epicycloid. — The  path  described  by  a  point 


Fig.  106. 


204 


ANALYTIC   GEOMETRY 


[Ch.  XV,  §  108 


on  the  circumference  of  a  circle  which  rolls  on  the  out- 
side of  a  fixed  circle  is  called  an  epicycloid. 

Let  the  student  show  that  the  equations  of  the  epi- 
cycloid are 

'a  +  bs 


x  —  (a-\-b~)  cos  (j>  —  b  cos 
y  =  («  +  b)  sin  <\>  —  b  sin 


b 

a  +  b 


* 


108.    The  cardioid.  —  An  important  special  case  of  the 
epicycloid  is  the  cardioid,  in  which  a  =  b;  but  instead  of 


Fig.  107. 


obtaining  its  equation  as  a  special  case  of  that  curve, 
we  shall  find  it  easier  to  obtain  its  polar  equation  at  once 
from  the  definition. 

Let  0,  the  original  point  of  contact  of  the  two  circles, 


Ch.  XV,  §  110]  HIGHER    PLANE    CURVES  205 

be  chosen  as  the  polar  origin,  and  the  diameter  CA  con- 
tinued through  this  point  as  the  polar  axis.  Let  the 
second  circle  roll  on  the  first  until  the  line  of  centres  CC 
makes  an  angle  0G7O"(=  </>)  with  its  original  position, 
and  let  P  be  the  new  position  of  the  point  of  contact. 
Then  OP  =  p,  and  the  angle  AOP  =  0.  From  0  and  P 
drop  the  perpendiculars  OM  and  PN  o\\  CC.  Evidently 
the  arc  OK  equals  the  arc  PK,  and  the  angles  PC C  and 
OCC  are  therefore  equal.  Then  OM=PN,  and  OP  is 
parallel  to  CC.     Then  6  =  <£. 

Now  CC  =  CM  +  MN  +  NC  =  2  a, 

or  2  CM+  OP  =  2  a, 

or  2  a  cos  6  -f  p  =  2  a, 

or  jo  =  2  a  (1  -  cos  0). 

109.  The  catenary.  —  The  curve  assumed  by  a  perfectly 
flexible  chain  of  uniform  weight  per  linear  unit,  when 
suspended  at  its  ends,  is  called  a  catenary.  Its  equation 
may  be  obtained  in  the  form 

where  e  is  the  base  of  the  Naperian  system  of  logarithms, 
and  the  origin  of  coordinates  is  a  units  below  the  lowest 
point  of  the  curve. 

110.  The  spirals.  —  The  curve  traced  by  a  point  which 
revolves  about  a  fixed  point,  and,  at  the  same  time,  re- 
cedes from  or  approaches  this  point  according  to  some 
definite  law,  is  called  a  spiral.      The  fixed  point  is  called 


206 


ANALYTIC   GEOMETRY 


[Ch.  XV,  §  110 


the  centre,  and  the  curve  traced  daring  one  revolution 
is  called  a  spire. 

If  any  two  radii  vectores  have  the  same  ratio  as  the 
angles  they  make  with  the  initial  line,  the  equation  of 
the  spiral  is  evidently  p  =  kO.  The  form  of  the  curve 
is  shown   in   Fig.   108.      The    dotted   line   indicates  the 

portion  of  the  locus  obtained  by 
giving  0  negative  values. 

Let  the  student  plot  the  spirals 
whose  equations  are 

The  curve  whose  equation  is 
log  p  =  Jc0,  or  p  =  a9,  is  called 
the  logarithmic  spiral.  When 
0  =  0,  p  =  1.  For  increasing 
positive  values  of  0,  p  increases 
very  rapidly ;  while  for  decreas- 
ing negative  values  of  0,  p  decreases  more  and  more 
slowly,  and  approaches  zero  as  a  limit.  There  are,  there- 
fore, an  indefinite  number  of  spires,  growing  smaller  as 
they  wind  about  the  origin,  but  never  passing  through 
that  point. 


Fig.  108. 


PART   II 

ANALYTIC   GEOMETRY   OF  SPACE 
CHAPTER  I 

COORDINATE  SYSTEMS.     THE  POINT 

1.  In  the  following  chapters  on  Analytic  Geometry  of 
Space,  a  knowledge  of  the  methods  and  results  of  Solid 
Geometry  and  of  Plane  Analytic  Geometry  is  presumed. 
Many  of  the  methods  and  formulas  to  be  given  for  three 
dimensions  are  closely  analogous  to  methods  and  formulas 
in  two  dimensions,  with  which  the  student  is  already 
familiar ;  and  in  all  such  cases  the  discussion  will  be 
condensed  into  as  brief  a  form  as  possible. 

For  convenience  of  reference,  the  following  theorems 
and  definitions  from  solid  geometry  are  cited  : 

If  a  straight  line  is  perpendicular  to  a  plane,  it  is  per- 
pendicular to  every  line  through  its  foot  in  the  plane. 

If  a  straight  line  is  perpendicular  to  any  two  straight 
lines  through  its  foot  in  a  plane,  it  is  perpendicular  to  the 
plane. 

The  angle  between  two  lines  not  in  the  same  plane 
is  the  same  as  the  angle  between  two  intersecting  lines 
parallel  respectively  to  the  given  lines. 

207 


208  ANALYTIC   GEOMETRY   OF   SPACE  [Ch.  1,  §  2 

The  orthogonal  projection  of  a  point  on  a  plane  (o:*  an 
axis)  is  the  foot  of  the  perpendicular  from  the  point  to 
the  plane  (or  the  axis).  The  projection  of  a  portion 
of  a  line  or  curve  on  a  plane  (or  an  axis)  is  the  locus 
of  the  projections  of  all  its  points. 

The  angle  which  a  line  makes  with  a  plane  is  the  angle 
which  it  makes  with  its  projection  on  the  plane. 

The  angle  between  two  planes  is  measured  by  the  angle 
between  two  lines,  one  in  each  plane,  drawn  perpendicular 
to  their  intersection  at  the  same  point. 

2.  Rectangular  coordinates.  —  In  applying  algebra  to 
the  geometry  of  space,  we  must  first  devise  some  method 
of  representing  the  position  of  a  point  in  space  by 
numbers. 

Construct  three  mutually  perpendicular  planes,  X-Y, 
Y-Z,  and  Z-X,  dividing  all  space  into  eight  compart- 
ments, called  octants.  These  planes  are  spoken  of  as 
coordinate  planes,  their  point  of  intersection,  0,  as  the 
origin,  and  their  lines  of  intersection,  OX,  OY,  and  OZ, 
as  coordinate  axes. 

A  point  in  space  is  located  by  means  of  its  distances, 
AP,  BP,  and  CP,  from  the  coordinate  planes,  measured 
parallel  to  the  coordinate  axes.  The  three  numbers  which 
represent  these  distances  are  called  the  rectangular  coor- 
dinates of  the  point,  and  are  always  written  in  the  order 

O*  y,  z). 

We  shall  consider  distances  as  positive  when  measured 
to  the  right,  forward,  or  upward ;  that  is,  parallel  to 
OX,  OY,  and  OZ.  Distances  measured  in  the  opposite 
directions  will  then  be  negative,     The  octant  0-XYZ  is 


Ch.  1,  §  3]        COORDINATE   SYSTEMS.     THE   POINT 


209 


V 


called  the  first,  and  the  others  may  be  numbered  in  any 
convenient  way. 

The  position  of  any  point  (#,  y,  z)  may  be  determined 
by  taking  on  the  axes  the  distances  OL,  OM,  and  6W, 
equal  to  these  coordinates, 
and  through  the  points  X, 
M,  iV,  passing  planes  parallel 
to  the  coordinate  planes, 
forming  a  rectangular  par- 
allelopiped ;  the  point  of 
intersection  of  these  planes  _ 
will  be  the  point  required. 

It  is  evident  that  rectan- 
gular coordinates  in  a  plane 
is  a  special  case  of  this  more 
general  system,  in  which  one 

of  the  coordinates  has  become  zero.  We  ought  therefore 
to  be  able  to  reduce  all  of  the  formulas  in  three  dimensions 
to  the  corresponding  formulas  in  two  dimensions  by  plac- 
ing z  equal  to  zero. 


-p<' 

i 
— i— 


>  i 

j  i 

J' 

0 


Fig.  1. 


PROBLEMS 
1.    Plot  the  following  points : 

(5,  4,  3),  (-  3,  4,  1),  (-  3,  -  1,  2),  (2,  -  3,  1),  (1,  1,  -  2), 
(-1,  4,  -2),  (-3,  -2,  -1),  (4,  -1,  -2);  (3,  4,  0), 
(-2,  0,  1),   (0,  -1,  3);  (5,  0,  0),  (0,  3,  0),    (0,  0,   -2). 

3.  Distance  between  two  points.  —  Let  Px  and  P2  be 
any  two  points  in  space,  and  through  each  of  them 
pass  three  planes  parallel  to  the  coordinate  planes,  form- 
ing a  rectangular  parallelopiped. 


210 


ANALYTIC   GEOMETRY  OF  SPACE  [Ch.  I,  §4 


Since    the    square    of    the   diagonal   of    a   rectangular 

parallelopiped  equals  the 
sum  of  the  squares  of  its 
edges, 


iyy  =  />,*,' +  JW 


+P1T1'. 

But  P1B1  =  x2  —  xv 

A#i  =  #2  -  VV 

and       P\TX  =  z9  —  zv 
Fig.  2. 

Hence      P1P2  =  V(*2  -  #i)2  +  (2/2  -  2/O2  +  (s2  -  »i)2.         [1] 
The    distance,   /a,    of    any    point    from    the    origin    is 

P  =  V^  +  2/2  +  z2.  [2] 


evidently 


4.    To  divide  a  line  in  any  given  ratio.  —  Let  the  point 

P  P      m 
P  divide  the  line  PXP2  so  that  —J—  = 


PP 


Project  the  line  PXP2 
on  the  X-  ^-plane,  form- 
ing the  trapezoid  P1C2, 
in  which  01P1  =  zv 
C2P2  =  z2,  and  OP  =  z. 
It  will  be  noticed  that 
this  is  the  same  figure 
used  in  Art.  13,  Part  I. 

Hence 


z- 


iniZ\  +  m,\Zi 


Ch.  I,  §5]        COORDINATE   SYSTEMS.     THE  POINT  211 

If  PxP<i  is  projected  un  the  other  planes,  we  obtain  in 

like  manner 

=  tggr,  +  mM          d       =  rn^  +  rnM .  rg-, 

ini  +  m*     *           "        rni  +  m?  L  J 

If  the  line  is  bisected,  these  formulas  become 

«=*4*.  V=^tU\  and   .  =  *  +  *  [4] 


2 


PROBLEMS 


1.  Find  the  length  of  the  line  joining  the  two  points 
(3,  2,  —  1)  and  (4,  —  2,  G)  and  the  coordinates  of  the  point 
which  divide  this  line  in  the  ratio  3  :  —  2. 

2.  Find  the  coordinates  of  the  centre  of  gravity  of  the  tri- 
angle whose  vertices  are  (.r„  yx,  z^),  (x.2j  y.2,  z2),  and  (x3,  y.3,  %). 

3.  Prove  that  in  any  tetraedron  the  four  lines  joining  the 
vertices  with  the  centres  of  gravity  of  the  opposite  faces  meet  in 
a  point,  which  is  three-fourths  of  the  distance  from  each  vertex 
to  the  opposite  face.  (This  point  is  the  centre  of  gravity  of 
the  tetraedron.) 

4.  Show  that  the  centre  of  gravity  of  any  tetraedron  bisects 
each  of  the  four  lines  joining  the  middle  points  of  the  opposite 
edges. 

5.  Show  that  the  straight  lines  which  join  the  middle  points 
of  the  opposite  sides  of  any  quadrilateral  meet  in  a  point  and 
are  bisected  at  that  point. 

6.  Show  that  the  sum  of  the  squares  of  the  diagonals  of  any 
quadrilateral  is  twice  the  sum  of  the  squares  of  the  lines  which 
join  the  middle  points  of  the  opposite  sides. 

5.  Projection  of  a  given  line  on  a  given  axis.  —  It  is 
required  to  find  the  projection  on  the  axis  OX  of  the  lind 
AB  which  makes  an  anode  a  with  the  axis. 


212 


ANALYTIC    GEOMETRY   OF   SPACE 


[Ch.  I,  §  6 


Through  A  and  B  pass  planes  perpendicular  to  the  axis, 
cutting  it  at  A'  and  B' .  Then  A'B'  will  be  the  pro- 
jection of  AB  on   OX.      Through  A  draw  the  line  AC 

parallel  to  OX.  Then 
A  C  =  A'B',  and  the  angle 
CAB=a.  In  the  right 
triangle  ABC, 

AC 

—X  • =  cos  a. 

AB 

Hence 

A'B'  =  AB  cos  a.     [5] 


Tig.  4. 


That  is,  the  projection  of  a  line  on  an  axis  is  equal  to  the 
length  of  the  line  multiplied  by  the  cosine  of  the  angle  which 
the  line  makes  with  the  axis. 

The  projection,  A'B',  of  a  directed  line,  AB,  is  evidently 
a  directed  line.  If  a  broken  line  AB,  BC,  CD  is  pro- 
jected on  an  axis,  the  algebraic  sum  of  the  projections 
of  its  parts  will  be  the  distance  along  the  axis  from  the 
projection  of  A  to  the  projection  of  D.  The  projection  on 
any  axis,  then,  of  any  closed  path  AB C...A  in  space,  which 
is  looked  upon  as  generated  by  the  movement  of  a  point 
from  A  to  B,  B  to  C,  etc.,  is  zero. 

6.  Polar  coordinates.  —  Let  OX,  OY,  OZ  be  a  set  of 
rectangular  axes  in  space,  and  let  P  be  any  point.  Draw 
OP. 

The  position  of  P  is  evidently  determined  if  we  know 
its  distance  p  from  the  origin,  and  the  angles  a,  /3,  and  7 
which  OP  makes  with  the  coordinate  axes. 

The  distance  p  is  called  the  radius  vector  of  the  point 


Ch.  I,  §  6]  COORDINATE    SYSTEMS.     THE    POINT 


213 


P ;  «,  /3,  and  7,  the  direction  angles  of  the  line  OP ;  and 
the  four  quantities  (jo,  «,  /3,  7),  the  polar  coordinates  of  the 
point.      Cos  a,  cos  /3,  and   cos  7  are   called   the   direction 


Fig.  5 


cosines  of  the  line  OP,  and  may  be  represented  by  the 
letters  Z,  ???.,  and  n. 

Let  L,  M,  and  JV  be  the  projections  of  P  on  the  axes. 
Then  OL  =  x,  031=7/  and  ON=z;  and  from  right  tri- 
angles we  have 

X  =  p  COS  a, 

y  =  p  COS  P,  [6] 

»  =  p  COS  Y. 

These  equations  give  the  relations  between  the  rec- 
tangular and  polar  coordinates  of  any  point.  Squaring 
and  adding,  we  have 

x2  +  y2  -f  22  =  p2  (cos2  a  +  cos2  ft  +  cos2  7) . 

But  from  [2],         p2  =  x2  +  y2  +  z2. 

Hence  cos2  a  +  cos2  p  +  cos2  y  =  !•  [7] 


214  ANALYTIC    GEOMETRY   OF   SPACE  [Ch.  I,  §  6 

That  is,  the  sum  of  the  squares  of  the  direction  cosines 
of  any  line  is  unity.  Hence  the  four  quantities  used  as 
polar  coordinates  of  a  point  are  equivalent  to  only  three 
independent  conditions,  as  we  should  expect. 

We  may  always  choose  these  coordinates  so  that  they 
shall  all  be  positive  and  so  that  the  angles  «,  /3,  y  shall 
not  be  greater  than  180°. 

since  any  line  parallel  to  OP  makes  the  same  angles 
with  the  axes,  we  may  define  the  direction  cosines  of  any 
line  in  space  as  the  same  as  the  direction  cosines  of  a  paral- 
lel through  the  origin. 

PROBLEMS 

1.  Find  the  direction  angles  of  a  line  equally  inclined  to 
the  three  axes. 

2.  If  I,  m,  and  n  are  the  direction  cosines  of  a  line,  show 
that  —  I,  —  m,  and  —  n  are  the  direction  cosines  of  the  same 
line  running  in  the  opposite  direction. 

3.  Find  the  direction  cosines  of  the  line  joining  the  origin 
to  the  point  (2,  6,  2),  and  the  projection  of  the  line  on  each  of 
the  coordinate  axes. 

4.  Find  the  direction  cosines  of  the  line  joining  the  points 
(2,  5,  1)  and  (3,  1,  8),  and  the  projection  of  the  line  on  each  of 
the  coordinate  axes. 

5.  Show  that  any  three  numbers  are  proportional  to  the 
direction  cosines  of  some  line. 

6.  Find  the  direction  cosines  of  a  line  which  are  propor- 
tional to  the  numbers  1,  2,  3. 

7.  Show  that  the  square  of  the  distance  between  two  points 
whose  polar  coordinates  are  (ply  «1?  /?2,  yx)  and  (p2,  ci2,  /?2,  72)  is 

p\  +  P22  —  2  pip2  (cos  %  cos  «,  +  cos  /?!  cos  (3.2  +  cos  yx  cos  y2). 

8.  A  line  makes  an  angle  of  60°  with  the  X-axis,  and  45D 
with  the  y-axis.     What  angle  does  it  make  with  the  Z-axis  ? 


Ch.  I,  §7]         COORDINATE   SYSTEMS.     THE   POINT 


215 


7.  Spherical  coordinates.  —  Let  OX,  OY,  OZ  be  a  set  of 
rectangular  axes  in  space,  and  let  P  be  any  point.  Draw 
OP,  and  pass  a  plane  through  OZ  and  OP.  The  position 
of  any  point  in  space  is  determined,  if  we  know  the  dis- 


Fig.  6. 


tance  p  from  the  origin  to  the  point  ;  the  angle  0  which 
the  plane  ZOP  makes  with  the  fixed  plane  ZOX;  and  the 
angle  <p  which  OP  makes  with  OZ. 

The  line  OZ  is  called  the  polar  axis,  and  the  point  0  the 
pole.  About  0  as  a  centre  describe  a  sphere  with  OP  as 
radius.  The  plane  ZOP  will  intersect  the  sphere  in  a 
meridian  circle.  The  angle  6  may  be  called  the  longitude 
of  P,  and  the  angle  <£,  the  colatitude.  The  distance  p  is 
called  the  radius  vector  and  (jo,  </>,  0)  are  called  the  spherical 
coordinates  of  P.  The  arrows  indicate  the  usual  choice  of 
positive  direction. 


216 


ANALYTIC    GEOMETRY   OF   SPACE 


[Ch.  I,  §  8 


Let  the  student  show  that  the  relations  between  rec- 
tangular and  spherical  coordinates  are 
oc  =  p  sin  <|>  cos  6, 

y  =  p  sin  <f>  sin  0,  [8] 

z  —  p  cos  <|>. 

Note.  —  Spherical  coordinates  have  usually  been  called  polar  coordi- 
nates. But  the  application  of  the  system  described  in  Art.  6  is  more 
nearly  analogous  to  the  uses  of  polar  coordinates  in  two  dimensions. 

8.  Angle  between  two  lines.  —  Let  uv  /3r  7r  and  «2,  /32,  y2 
be  the  direction  angles  of  two  lines,  and  let  6  be  the  angle 
between  them.  Draw  parallels  to  these  lines  through  the 
origin,  and  on  each  of  these  parallels  take  a  point,  as  Px 
and  P0. 


Then  by  [1] 


Fig.  7. 


*VY  =  Oi  -  ^)2  +  (jfi  -  y*y  +  Oi  -  *2)2> 

or  by  [6]     =  (px  cos  ax  —  p2  cos  «2)2~l~G0i  cos  ^i~p2  cos  ^)2 

+  0>icos71-/?2cos72)2, 
or  by  [7]     =  Pi+p^   —  2  pxp2  (cos  ax  cos  a  2  +  cos  /3X  cos  fi2 

-f-  cosyj  cos72). 


Ch.  I,  §  8]        COORDINATE   SYSTEMS.     THE   POINT  217 

But  by  the  law  of  the  cosines 

Hence  cos  0  =  cos  ax  cos  a2  +  cos  pi  cos  p2  +  cos  71  cos 72.        [9] 

If  the  lines  are  perpendicular  cos  6  =  0,  and  the  condi- 
tion for  perpendicularity  is 

cos  ai  cos  a2  4-  cos  Pi  cos  p3  +  cos  -yi  cos  Y2  =  0.  [10] 

If  the  lines  are  parallel,  they  must  make  the  same  angles 
with  the  axes,  and  the  conditions  for  parallelism  are 

01  =o2,  Pi  =  p2,  and  71  =  Y2.  [11] 

PROBLEMS 

1.  Show  that  the  three  lines  whose  direction  cosines  are 

12-3-4.       4        12        3    .     orirl       3        -4       1  2 
T5>  T^>    T 3  J    T¥>   T 3 >    T3  J     djllu   T3 J    T 3 >   T 3 

are  mutually  perpendicular. 

2.  Show  that  (3,  30°,  60°,  90°),  and  (5,  30°,  90°,  60°)  are 
possible  polar  coordinates  of  two  points,  and  find  the  angle  they 
subtend  at  the  origin. 

3.  Show  that  the  conditions  for  parallelism  are  consistent 
with  [9]  when  0  =  0°. 

4.  Find  the  rectangular  coordinates  of  the  points  in 
problem  2. 

5.  Find  the  polar  coordinates  of  the  point  (3,  —  6,  2). . 

6.  Find  the  angle  subtended  at  the  point  (1,  2,  3)  by  the 
points  (2,  3,  4)  and  (5,  4,  3). 


218 


ANALYTIC   GEOMETRY   OF   SPACE  [Ch.  I,  §  10 


9.    Transformation  of  coordinates.     Parallel  axes.  —  If 
the  new  axes  are  parallel  to  the  old,  and  the  coordinates 

of  the  new  origin,  re- 
ferred to  the  old  axes,  are 
(#0,  y0,  20),  the  equations 
of  transformation  are 
easily  seen  (see  Fig.   8) 


x  to  be 


ac  =  x0  +  oc'f 

V  =  2/o  +  V,     [12] 

Z  =  Z0  +  Z  . 


10.  Transformation  of  coordinates  from  one  set  of  rec- 
tangular axes  to  another  which  has  the  same  origin.  —  Let 
(«ii  fiv  Yj),  («2,  £2,  72),  and  («3,  /33,  y3)  be  the  direction 
angles  of  OX\  OY\  and 
OZ'  with  respect  to  the 
original  axes.  The  coor- 
dinates (#,  y,  2)  of  any 
point  P  are  the  projec- 
tions of  OP  on  OX,  OF, 
and  OZ.  But  the  broken 
line  made  up  of  x\  y!,  and 
z'  extends  from  0  to  P, 
and  will  therefore  have 
the  same  projections  on  Y 
the  axes  as  OP.  Hence 
(by  Art.  5) 


Fig.  9. 


ac  =  x'  COS  ai  +  y'  cos  a2  +  Z'  COS  a3, 
y  =  05'  cos  Pi  +  y'  cos  p2  +  s'  cos  p3, 
3  =  05'  cos  71  +  2/'  cos  72  +  z'  COS  73. 


[13] 


Ch.  I,  §  10]      COORDINATE   SYSTEMS.     THE   POINT  219 

Let  the  student  show  that  the  transformation  of  coordi- 
nates cannot  alter  the  degree  of  an  equation.  (See  Art. 
50   Part  I.) 

PROBLEMS 

1.  What  will  be  the  direction  cosines  of  OX,  OY,  and  OZ 
referred  to  the  new  axes  in  Art.  10  ? 

2.  What  six  relations  hold  between  alf  ft,  y1}  u2,  ft,  etc., 
from  [7]  ? 

3.  What  six  relations  hold  between  u1}  ft,  yl9  a2,  (32,  etc., 
from  [10]  ? 

4.  Show  that  the  twelve  relations  obtained  in  problems  2 
and  3  are  equivalent  to  only  six  independent  conditions.  How 
many  of  the  coefficients  in  equations  [13]  are  independent  ? 


CHAPTER   II 
LOCI 

11.  Equation  of  a  locus. — If  a  point  moves  in  space 
according  to  some  law,  it  will  generate  some  locus.  As, 
for  example,  a  point  keeping  at  a  fixed  distance  from  a 
fixed  point  will  generate  the  surface  of  a  sphere.  If  we 
can  translate  the  statement  of  the  law  into  an  algebraic 
relation  between  the  coordinates  of  the  points  which  satisfy 
the  law,  we  shall  have,  as  in  plane  analytic  geometry,  an 
equation  which  can  be  used  to  represent  the  locus.  In 
the  above  example,  if  the  origin  is  at  the  centre,  the  equa- 
tion of  the  surface  will  be  x2  +  y2  +  z2  =  r2 ;  for  this  states 
that  the  point  (x,  y,  z),  which  satisfies  it,  must  remain  at 
the  distance  r  from  the  origin. 

The  planes  parallel  to  the  coordinate  planes  are  evi- 
dently represented  by  x  =  kv  y  =  &2,  and  z  =  Jcs ;  for  these 
equations  state  that  the  points  which  satisfy  them  are  at  a 
fixed  distance  from  the  coordinate  planes. 

PROBLEMS 

1 .  What  are  the  equations  of  the  coordinate  planes  ? 

2.  What  are  the  equations  of  the  planes  bisecting  the  angles 
between  the  X-Y  and  5 -Z-planes  ?  Between  the  Y-Z  and 
Z-X-planes  ? 

3.  What  equation  must  be  satisfied  by  the  coordinates  of  a 
point  which  remains  at  a  distance  of  5  units  from  the  X-axis  ? 
5  units  from  the  F-axis  ?     What  is  the  locus  in  each  case  ? 

220 


Ch.  II,  §§  12,  13]  LOCI  221 

4.  Find  the  equation  of  the  locus  of  a  point  which  is  5 
units  from  the  point  (3,  2,  5). 

5.  What  equations  must  be  satisfied  by  the  coordinates  of 
a  point  which  is  equidistant  from  the  three  points  (1,  3,  8), 
(-  6,  -  4,  2),  and  (3,  2,  1)  ? 

12.  Cylindrical  surfaces.  —  If  a  cylindrical  surface  is 
formed  by  the  movement  of  a  line,  which  remains  parallel 
to  one  of  the  axes,  while  moving  along  a  directing  curve 
in  the  plane  of  the  remaining  axes,  its  equation  in  three 
dimensions  will  be  the  same  as  the  equation  in  two  dimen- 
sions of  the  directing  curve,  and  will  contain  only  two 
variables.  For,  suppose  the  line  remains  parallel  to  the 
Z-axis  and  the  directing  curve  lies  in  the  X-Z-plane  ;  then, 
for  any  position  of  the  line,  the  relation  between  the  x  and 
y  coordinates  of  any  point  on  it  will  be  the  same  as  the 
relation  between  the  x  and  y  coordinates  of  the  point  where 
the  line  touches  the  directing  curve,  while  the  z  coordi- 
nate may  have  any  value  whatever.  The  equation  in  x 
and  y  of  the  directing  curve  is,  therefore,  the  only  necessary 
relation  between  the  coordinates  of  any  point  on  the  sur- 
face, and  as  it  is  not  satisfied  by  any  point  not  on  the 
surface,  it  is  (when  interpreted  as  an  equation  in  three 
dimensions)  the  equation  of  the  surface. 

In  a  similar  manner,  it  may  be  shown  that  the  equations 
of  cylindrical  surfaces,  whose  elements  are  parallel  to  the 
JT-axis,  contain  only  y  and  z  ;  parallel  to  the  !F-axis,  only 
x  and  z. 

13.  Surfaces  of  revolution.  —  Surfaces  generated  by  the 
revolution  of  a  plane  curve  about  one  of  the  coordinate 
axes  form  another  class  of  surfaces  whose  equations  can 
be  determined  easily. 


222 


ANALYTIC   GEOMETRY   OF   SPACE       [Ch.  II,  §  13 


For  example,  let  it  be  required  to  determine  the  equation 
of  the  surface  generated  by  the  revolution  of  the  ellipse 

1  about  the  X-axis.      Let  P'  (V,  y\  z')  be  any 
z 


a?    r 

a2_h62 


Fig.  10. 


point  on  the  surface,  and  through  P'  pass  a  plane  perpen- 
dicular to  the  X-axis.  The  section  of  the  surface  made 
by  this  plane  is  evidently  a  circle.      Hence  LP'  =  LK. 

But  LP'  =  V/M^T2"  and  OL  =  x\ 

The  coordinates  of  iT  in  the  X-P-plane  are,  therefore, 
x'  and  Vy2  +  z'2,   and  since  K  is  a  point  on  the  ellipse 

1,  these  coordinates  must  satisfy  that  equation, 


^2      f 
a2_h62 


or 


+ 


y">  +  z> 


=  1. 


Dropping  primes,  we  have  as  the  equation  of  an  ellipsoid 
of  revolution  about  the  X-axis, 

a2"^2"^2" ' 


Ch.  II,  §  14]  LOCI  223 

A  general  rule  for  finding  the  equation  of  a  surface  of 
revolution,  formed  by  revolving  a  plane  curve  about  one 
of  the  coordinate  axes,  may  be  stated  thus:  Replace  in  the 
equation  of  the  plane  curve  the  coordinate  perpendicular  to 
the  axis  of  revolution  by  the  square  root  of  the  sum  of  the 
squares  of  itself  and  of  the  third  coordinate. 

PROBLEMS 

1.  Find  the  equation  of  the  surface  generated  by  a  line 
moving  parallel  to  the  Z-axis  along 

x2     if 

(a)  the  ellipse  -  +  |-2  =  1, 

(b)  the  parabola  y2  =  2  mx, 

(c)  the  line  x  +  3  y  =  6. 

2.  What  is  the  equation  of  a  circular  cylinder  whose  axis  is 
parallel  to  the  F-axis  and  passes  through  the  point  (3,  0,  5), 
and  whose  radius  is  5. 

3.  Find  the  equation  of  the  surface  of  revolution,  formed  by 
revolving  about  the  X-axis 

(a)  the  line  y  =  4,  (a  cylinder) 

(6)  the  line  x  =  y,  (a  cone) 

(c)  the  circle  x2  -\-  y2  =  r2,  (a  sphere) 

(cZ)  the  parabola  y2  =  2  mx,  (a  paraboloid  of  revolution). 

4.  Obtain  the  equations  of  the  hyperboloids  of  revolution 
formed  by  revolving  the  hyperbola  about  (a)  its  transverse 
axis ;  (b)  its  conjugate  axis. 

14.  Locus  of  an  equation. — Again,  as  in  plane  analytic 
geometry,  an  equation  between  x,  y,  and  z  expresses  a 
necessary  relation  between  the  coordinates  of  every  point 
which  satisfies  it,  and  hence  cannot  be  satisfied  by  points 
taken  at  random  in  space.      It  is  easy  to  see  that  the 


224  ANALYTIC   GEOMETRY   OF   SPACE       [Ch.  II,  §  14 

points  which  satisfy  it  may  be  taken  as  near  to  each  other 
as  we  please.  Moreover,  any  such  equation  represents  a 
surface  of  some  kind,  as  we  shall  now  prove. 

Let/(:r,  y,  z)=  Q  be  an  equation  of  any  degree  between 
x,  y,  and  z.  If  we  substitute  x  —  k  (any  constant),  the 
resulting  equation,  f(y,  z}  =  0,  must  represent  the  rela- 
tion between  y  and  z  for  all  points  of  the  locus  for  which 
x  =  k,  or  which  lie  in  a  plane  at  distance  k  from  the 
Y-Z-plane.  But  since  the  locus  of /(?/,  z)  =  0  lies  wholly 
in  this  plane,  it  is  a  plane  curve.  Hence  the  intersection 
of  any  plane  parallel  to  the  Y-Z-plnne  with  the  locus  of 
f(x,  y,  z)  =  0  is  a  plane  curve.  This  can  be  proved  in 
like  manner  for  all  planes  parallel  to  the  X-Y  and  X-Z- 
planes.  If  the  axes  are  revolved  through  any  angle,  the 
equation  of  the  locus  will  be  of  the  same  general  form 
and  every  plane  parallel  to  the  new  axes  will  cut  it  in  a 
plane  curve.  Hence  all  planes  cut  the  locus  in  a  plane 
curve,  and  the  locus  is  therefore  a  surface. 

If,  in  particular,  the  equation  is  of  the  first  degree,  its 
intersection  with  any  of  these  planes  will  be  a  straight 
line.  An  equation  of  the  first  degree  therefore  always 
represents  a  plane. 

If  an  equation  does  not  contain  a  term  in  z,  the  relation 
between  x  and  y  will  not  be  changed  by  a  change  in  z. 
The  sections  of  the  locus  parallel  to  the  X-F-plane  are 
therefore  all  alike,  and  the  locus  is  a  cylindrical  surface, 
having  all  its  elements  parallel  to  the  Z-axis.  In  like 
manner,  if  an  equation  does  not  contain  a  term  in  y,  it 
represents  a  cylindrical  surface  parallel  to  the  y-axis; 
if  it  contains  no  term  in  x,  a  cylindrical  surface  parallel 
to  the  X-axis. 


Ch.  II,  §  14]  LOCI  225 

If  in  particular  the  equation  is  of  the  first  degree,  the 
surface  becomes  a  plane  parallel  to  one  of  the  axes. 

If  two  equations  are  simultaneously  satisfied  by  the 
coordinates  of  points  on  a  locus,  that  locus  must  consist 
of  the  points  common  to  the  loci  of  the  two  equations. 
Hence  two  equations  of  the  form  fx  (#,  y,  z)  =  0  and 
f2 (x,  y,  z)  =  0,  taken  togetlier,  represent  a  curve  in  space, 
the  intersection  of  the  surfaces  which  they  represent. 

In  particular,  if  these  twTo  equations  are  of  the  first 
degree,  this  locus  will  be  the  intersection  of  the  two 
planes  which  they  represent.  Hence  two  equations  of  the 
first  degree,  used  simultaneously,  represent  a  straight  line. 

Three  equations  used  simultaneously  are  satisfied  by  the 
coordinates  of  a  finite  number  of  points  only,  —  the  points 
of  intersection  of  the  curve  represented  by  two  of  the  equa- 
tions with  the  surface  represented  by  the  third. 

The  curves  of  intersection  of  any  surface  with  the 
coordinate  planes  are  called  the  traces  of  the  surface. 
Their  equations  may  be  found  from  the  equation  of  the 
surface  by  placing  each  of  the  coordinates  in  turn  equal 
to  zero. 

The  general  method  of  determining  the  form  of  the 
surface  represented  by  any  given  equation  will  be  taken 
up  in  the  chapter  on  quadric  surfaces. 

PROBLEMS 

1.    What  surface  is  represented  by  the  equations  ? 
(a)  x  =  y,  (d)  x2  +  y2  =  25, 

(!>)  y  =  z,  (e)   x2  +  f-  +  z2=  25, 

(c)  x-y=5,  (f)x*-2y  =  0. 


226  ANALYTIC   GEOMETRY   OF   SPACE       [Ch.  II,  §  14 

2.  Obtain  the  traces  on  each  of  the  coordinate  planes  of  the 
loci  of  the  following  equations,  and  from  these  traces  deter- 
mine roughly  the  nature  of  the  surface: 

(a)  x?  +  tf  =  9,  (d)  y2  =  ±z, 

(b)  x-y  +  2z  =  10,  (e)   J  +  J  +  g^l, 

(c)a?  =  2y,  (f)x>  +  tf-2z  =  0. 

3.  What  is  the  equation  of  the  surface  generated  by  the 
revolution  of  the  hyperbola  xy  =  k  about  the  X-axis. 

4.  What  is  the  position  of  a  line  whose  equations  are 
x  +  3y  =  10  and  Sx  —  4,y  =  S? 

5.  The  equations  of  any  two  surfaces  may  be  represented 
by  U=  0  and  V=  0,  where  U  and  Fare  abbreviations  for 
algebraic  expressions  of  any  degree  in  x,  y,  and  z.  Show 
that  lU+kV=0  will  represent  a  surface  which  passes 
through  all  the  points  common  to  the  loci  of  U  =  0  and 
V=  0,  and  which  meets  neither  of  these  surfaces  at  any 
other  points.  Show  also  that  the  locus  of  UV=  0  will 
consist  of  the  loci  of  TJ—  0  and   V=  0. 


CHAPTER   III 


THE  PLANE 

15.  Normal  form  of  the  equation  of  a  plane. — Let  ON 
be  the  normal  to  the  plane  (a  straight  line  of  indefinite 
extent  perpendicular  to  the  plane),  and  let  <*,  /3,  and  7  be 
the  angles  which  this  normal  makes  with  the  axes.  Let 
p  be  the  perpendicular  distance  OK  from  the  origin  to 


Fig.  11. 


the  plane,  measured  along  the  normal.  Let  P(x,y,z) 
be  any  point  in  the  plane.  The  line  PK  will  be  perpen- 
dicular to  ON,  and  the  projection  of  OP  on  ON  will  be 
OK  or  p.  But  the  projection  of  OP  on  ON  is  the  same 
as  the  projection  on  ON  of  the  broken  line  OL,  LC,  CP, 

227 


228  ANALYTIC   GEOMETRY   OF   SPACE       [Ch.  Ill,  §  16 

or  #,   ?/,  z.     From  [5]  the   projection    of  OL  on  OJV  is 
#cos«;    of  L  C,  y  cos  /3 ;    of  CP,  2  cos  7. 

Hence  oc  cos  a  -f  y  cos  p  +  2  cos  y  -  p  =  0.  [14] 

This  is  called  the  normal  form  of  the  equation  of  a 
plane. 

The  distance  p  is  measured  from  the  origin  to  the 
plane,  and  is  positive  or  negative  according  as  it  runs 
in  the  positive  or  negative  direction  of  the  normal.  It 
is  usually  possible  to  choose  the  direction  from  the  origin 
to  the  plane  as  the  positive  direction  of  the  normal,  so 
that  p  will  usually  be  a  positive  number. 

The  angles  «,  /3,  and  7  are  measured  from  the  positive 
directions  of  the  axes  to  the  positive  direction  of  the 
normal. 

16.  Reduction  of  the  general  equation  Ax  +  By  +  Cz 
+  D  =  0  to  the  normal  form.  — It  lias  been  shown  in  the 
previous  chapter  that  every  equation  of  the  first  degree 
represents  a  plane.  Let  the  general  equation  of  the  first 
degree,  Ax  +  By  +  Cz  +  D  =  0,  be  the  equation  of  a  plane, 
and  let  x  cos  a  +  y  cos  /3  -f  z  cos  7  —  p  =  0  be  the  equation 
of  the  same  plane  in  the  normal  form.  Then,  since  the 
two  equations  represent  the  same  plane,  they  can  differ 
only  by  a  common  factor.     Then  JcA  =  cos  a,  kB  =  cos  /3, 

and  JcC=  cos 7.      Hence   h  = ,    and   the 

±  V  J2  +  B2  +  O2 
equation  ^       T 


a?  +  ' —  1/  + 


+      ,      n      =  =  0  [151 


Ch.  Ill,  §  18]  THE    PLANE  229 

is  in  the  normal  form.  If  we  wish  to  keep  p  positive,  it 
is  necessary  to  choose  the  sign  of  the. radical  opposite  to 
the  sign  of  D.     Then  the  coefficient  of  x  is  cos  «,  etc. 

17.    Equation  of  a  plane  in  terms  of  its  intercepts.  —  If 

the  intercepts  of  a  plane  on  the  axes  are  a,  5,  and  c,  the 
coordinates  of  the  points  where  it  cuts  the  axes  are 
(a,  0,  0),  (0,  5,  0),  and  (0,  0,  c).  If  these  coordinates 
are  substituted  successively  in  the  general  equation 

Ax  +  By  +  Cz  +  D  =  0, 

we  have       a  = -,    6  =  — -,    and    c  = -• 

A  B  V 

But  the  general  equation  may  be  written  in  the  form 

x  y  z         _, 

+ 7T=L 


A  J5  (7 

From  this  we  have,  by  substitution, 

°°  +  V  +  s  =  l  [16] 

a     b     c  L      J 

as  the  equation  of  a  plane  in  terms  of  its  intercepts. 

18.  Distance  of  a  point  from  a  plane.  —  Let  it  be  re- 
quired to  find  the  distance  of  the  point  Px  from  the  plane 
UK,  when  the  equation  of  UK  is  given  in  the  form 

x  cos  «  4-  y  cos  /3  -f-  z  cos  7  —  p  =  0. 

Pass  a  plane  RS  through  Pv  parallel  to  UK.  Its 
equation  will  be 

x  cos  a  +  y  cos  /3  -f-  2  cos  7  —  jt^  =  0, 


Fig.  12. 


230  ANALYTIC   GEOMETRY   OF   SPACE     [Ch.  Ill,  §  18 

where    px   can   be    either   positive  or   negative,  since   it 

is  the  distance  from  the 
origin  to  the  plane  BS, 
measured  along  the  nor- 
mal to  UK.  The  coordi- 
nates of  Px  must  satisfy 
the  equation  of  BS. 
Hence 

xx  cos  a  -f  yx  cos  ft 

+  zx  cos  7  =  pv 

Now,  wherever  Px  may 
lie, 

MPX  =  NNX  =  ONx  -  ON=px  -p 

=  xx  cos  a  +  y1  cos  ft  +  z1  cos  7  —  p. 

If  the  equation  is  given  in  the  form 

Ax  +  By  +  Cz  +  D  =  0, 

MP,  =  Ax*  +  ^  +  Czi+M,  r171 

±  V^  +  B*  +  C2  L      J 

where  the  sign  of  the  radical,  is  chosen  opposite  to  that  of 
D.  The  distance  MP1  is  positive  when  the  point  and 
the  origin  are  on  opposite  sides  of  the  plane ;  negative 
when  they  are  on  the  same  side  of  the  plane. 

PROBLEMS 
1.    Given  the  plane  3x  —  Sy  -\-  z  =  12,  find 

(a)  the  direction  cosines  of  a  normal, 

(b)  its  distance  from  the  origin, 

(c)  its  distance  from  the  point  (3,  —  2,  6), 

(d)  its  intercepts. 


Ch.  Ill,  §  10]  THE    PLANE  231 

2.  Find  the  equation  of  a  plane,  if  the  foot  of  the  perpen- 
dicular from  the  origin  on  it  is  the  point  (3,  1,  —  5). 

3.  On  which  side  of  the  plane  7  x-\-Ay  =  5  is  the  point 
(0,  7,  3)  ?  How  is  this  plane  situated  ?  What  are  its  traces 
on  the  coordinate  planes  ? 

4.  Show  that  the  three  planes  2x  +  5y  -\-  3 2  =  0,  x  —  y 
+  42  =  2,  and  1  y  —  524-4  =  0  intersect  in  a  straight  line. 

5.  Find  the  equation  of  the  plane  which  bisects  the 
angle  between  the  two  planes  A^x  4-  B^  +  Cxz  +  Dx  =  0  and 
Afc  +  5^/4-  C#  4-  A  =  0. 

6.  Find  the  equation  of  a  plane  through  the  origin  and 
the  line  of  intersection  of  the  planes  x  4-  3  y  —  4  z  =  10  and 
5y— 624-3  =  0.     (See  problem  5,  page  214.) 

19.  The  angle  between  two  planes.  — The  angle  between 
two  planes  is  easily  seen  to  be  equal  to  the  angle  between 
their  normals. 

If  the  two  planes  are 

x  cos  «!  -f  y  cos  /3X  4-  z  cos  yx  —  p1  =  0, 
and  x  cos  «2  4-  y  cos  /32  +  2  cos  y2  —  /?2  =  0, 

the  angle  between  them  is  given  by 

cos  9  =  cos  ai  cos  a2  +  cos  Pi  cos  p2  4-  cos  71  cos  72.         [18] 
If  the  two  planes  are      Axx  4-  Bxy  4-  Cxz  +  Dl  =  0, 
and  J.2z  -f-  B2y  4-  <?22  4-  i>2  =  0, 

the  angle  between  them  is  given  by 

cose  = AlA*  +  BlB*  +  Cl^2  ["191 

±  V^  +  .B^  +   Cl2  '   V^22  4"  ^22  +    C22 

If  the  sign  of  the  first  radical  is  chosen  opposite  to  the 
sign  of  Dv  and  the  sign  of  the  second  opposite  to  the  sign 


232  ANALYTIC   GEOMETRY   OF   SrACE     [Ch.  Ill,  §  21 

of  D2,  6  will  be  the  angle  between  the  positive  directions 
of  normals  to  the  planes. 

20.   Perpendicular  and  parallel  planes. — If  two  planes 
are  perpendicular,  cos  6  =  0, 
and  AiA2  +  BiB2  +  dC2  =  0.  [20] 

If  two  planes  are  parallel,  the  direction  cosines  of  their 
normals  must  be  equal, 


or 


and 


4 

A 

V^2 

+  B? 

Br 

+  0* 

VA2 

+  A2 

B2 

+  c22 

V^!2 

+  B* 
Or 

+  c* 

V^2S 

0% 

'+<V 

V^2  +  B*  +  C\2      V^l22  +  B}  +  C22 
The  conditions  for  parallelism  are,  therefore, 


A1  =  B1=Clt 

A2     B2      Co 


[21] 


Notice  that  two  planes  will  be  perpendicular  when  a 
single  condition  is  satisfied;  but  that  two  conditions  must 
be  satisfied  if  the  two  planes  are  to  be  parallel. 

21.   The  equation  of  a  plane  satisfying  three  conditions.  — 

The  general  equation  of  a  plane,  Ax  +  By  +  Cz  +  D  =  0, 
contains  three  independent  coefficients,  and  therefore  three 
independent  relations  between  the  coefficients  will  deter- 
mine the  plane.  For  any  three  such  conditions  will  give 
three  equations  between  the  four  coefficients,  from  which 
three  of  the  coefficients  can  be  determined  in  terms  of  the 
fourth.  If  we  substitute  these  values  in  the  general 
equation,  and  divide  by  the  fourth  coefficient,  the  equa- 
tion is  completely  determined, 


Ch.  Ill,  §  21]  THE   PLANE  233 

The  coordinates  of  three  points  are  three  conditions 
from  which  three  such  equations  can  be  obtained;  for 
these  sets  of  coordinates  must  each  satisfy  the  general 
equation.  If  the  three  points  happen  to  lie  on  a  line,  the 
equations  for  determining  the  coefficients  will  not  be  inde- 
pendent, and  the  plane  will  not  be  determined. 

Again,  a  plane  can  be  determined  which  shall  pass 
through  two  points  and  also  be  perpendicular  to  a  given 
plane ;  for  the  substitution  of  the  coordinates  of  the  two 
points  will  give  two  equations  between  the  coefficients, 
and  the  condition  for  perpendicularity  [20]  will  give  a 
third.  If,  however,  the  two  points  lie  on  the  same  normal 
to  the  plane,  the  solution  will  be  indeterminate,  since  the 
conditions  will  not  be  independent.  Again,  a  plane  can  be 
determined  which  shall  pass  through  one  point  and  also 
be  parallel  to  a  given  plane  ;  for  the  substitution  of  the 
coordinates  of  the  point  gives  one  equation  between  the 
coefficients,  and  the  conditions  for  parallelism  [21]  give 
a  second  and  third. 

But  here  there  is  a  simpler  method  ;  if  the  equation  of 
the  plane  is  Ax  -f  By  +  Cz  +  D  =  0,  any  plane  parallel 
to  it  may  be  written  in  the  form  Ax  -{-By  +  Cz  +DX  =  0, 
since  the  conditions  for  parallelism  are  satisfied.  We  can 
determine  D1  from  the  fact  that  the  coordinates  of  the 
point  must  satisfy  the  equation. 

PROBLEMS 
1.    Find  the  equation  of  a  plane  through  the  points 

(a)  (4,2,1),  (-1,-2,2),  (0,4,-5), 

(b)  (-1,-1,-1),  (3,2,-2),  (2,0,0). 


234  ANALYTIC   GEOMETRY   OF   SPACE     [Ch.  Ill,  §  21 

Find  the  intercepts  of  these  planes  on  the  axes  and  their 
distances  from  the  origin. 

2.  Find  the  equation  of  a  plane  through  the  points  (2, 1,  —1) 
and  (1,  1,  2),  and  perpendicular  to  the  plane  7  #  -f-  4  ?/  —  4  z 
-36  =  0. 

3.  Find  the  equation  of  a  plane  through  the  points  (2,  0,  —1) 
and(l,  —6,1),  and  perpendicular  to  the  plane  ox+3y— z  —  4  =  0. 

4.  Find  the  equation  of  a  plane  through  the  point  (2, 1,  —1) 
and  parallel  to  the  plane  7  as  +  4  y  —  4  z  +  36  =  0. 

5.  Find  the  equation  of  a  plane  which  bisects  the  line  join- 
ing the  two  points  (6,  4, 1)  and  (2,  4,  —1),  and  is  perpendicular 
to  that  line. 

6.  Find  the  equation  of  a  plane  which  passes  through  the 
origin  and  is  perpendicular  to  the  two  planes  2x— 4?/-}-3  2=12 
and  7x  +  2y  +  z  =  0. 

7.  Prove  that  the  six  planes,  each  containing  one  edge  of  a 
tetraedron  and  bisecting  the  opposite  edge,  meet  in  a  point. 

Note. — The  coordinates  of  the  point  of  intersection  of  three  planes 
may  be  found  by  solving  the  three  equations  simultaneously. 

8.  Prove  that  the  six  planes,  each  passing  through  the  mid- 
dle point  of  one  edge  of  a  tetraedron  and  being  perpendicular 
to  the  opposite  edge,  meet  in  a  point. 


CHAPTER    IV 

THE   STRAIGHT   LINE 

22.    Equations.  —  We  have  seen  that,  if  a  point  moves 

in  space  in  such  a  way  as  to  satisfy  at  the  same  time  two 

equations  of  the  first  degree,  the  locus  which  is  generated 

is  the  line  of  intersection  of  their  planes.     Then  the  two 

equations 

Axx  +  Bxy  +  Cxz  +  2>i  =  0, 

and  A2x  +  B2y  +  C2z  +  D2  =  0 

will  in  general  represent  a  line,  the  only  exception  being 

when  — J  =  — 1  =  —J.,  and  the  planes  are  parallel. 
A2      B2      C2 

But  the  line  may  be  determined  by  any  pair  of  planes 
which  pass  through  it,  and  it  is  convenient  to  pick  out 
those  planes  which  have  the  simplest  form.  The  equa- 
tion of  any  plane  through  the  line  can  be  written  in  the 
form 
Axx  +  BlV  +  Cxz  +  J)1  +  k(A2x  +  B2y  +  02z  +  D2)=  0. 

When  none  of  the  coefficients  Av  Bv  etc.,  are  zero,  it 
will  always  be  possible  to  choose  k  in  such  a  way  as  to 
eliminate  y  and  reduce  the  equation  to  the  form  x  =  mz  +  a. 
Again,  k  may  be  so  chosen  as  to  eliminate  x  and  reduce 
the  equation  to  the  form  y  =  nz  -f-  b.     Then  the  equations 

x  =  mz  +  a, 
and  y  =  nz  -f  6, 

235 


236  ANALYTIC   GEOMETRY   OF   SPACE     [Ch.  IV,  §  22 

each  determine  a  plane  through  the  line,  and  hence  may 
be  used  as  the  equations  of  the  line.  These  planes  are 
seen  to  be  the  projecting  planes  of  the  line,  perpendicular 
to  the  X-Z  and  Y-Z-planes.  The  equations  of  any  two  of 
the  three  projecting  planes  may  be  chosen  as  the  equations 
of  the  line. 

In  practice,  to  reduce  the  equations  of  a  line  to  their 
simplest  form,  we  simply  eliminate  one  of  the  variables 
and  then  another  from  the  two  equations.  Indeed,  it  is 
evident  algebraically  that  any  set  of  values  which  satisfy 
a  pair  of  equations  must  also  satisfy  any  equation  which 
can  be  deduced  from  them. 

If  some  of  the  coefficients  Av  Bv  etc.,  are  zero,  it  will 
always  be  possible  by  elimination  to  reduce  the  equations 
to  one  of  the  three  forms 

x  =  mz  +  #>  y  =  qx  +  c,         x  =  e, 

or 
y  =  nz  +  5,  z  =  d,  z  =f. 

The  first  form  includes  all  lines  not  parallel  to  the  X-Y- 
plane  ;  the  second,  lines  parallel  to  the  X-!F-plane,  but 
not  parallel  to  the  y-axis ;  the  third,  lines  parallel  to  the 
P-axis. 

PROBLEMS 

1.  Write  the  equations  of  each  of  the  coordinate  axes. 

2.  Write  the  most  general  form  of  the  equations  of  a  line  in 
each  of  the  coordinate  planes ;  parallel  to  each  of  the  coordi- 
nate planes ;  parallel  to  each  of  the  coordinate  axes. 

3.  Show  how  to  find  the  points  where  a  given  line  pierces 
the  coordinate  planes,  and  by  this  means  plot  the  lines  in 
problem  4. 


Ch.  IV,  §  23] 


THE   STRAIGHT   LINE 


237 


4.    Reduce  these  equations  to  their  simplest  forms 

(a)  2a?-3#  +    z  -  6  =  0,  (6)   2x  +  3y 

x  +    7/  —  3  z  —  1  =  0. 
(c)   2a?  +  4y  +  3«  +  6  =  0, 

3a?  +  6y  +  22-l  =  0.  3t/ 

(e)  2a>-3y-    z  +  2  =  0,  (/)  4.y 

4a?-6y  +  32J-l  =  0.  2t/ 


6  z  -  12  =  0, 


4#  —     t/  +  12z  +    4  =  0. 
(rt)   4 y +  3*+    1=0, 


—  2 


12  =  0. 

3z-    2  =  0, 

2  +    4  =  0. 


5.  Find  the  equations  of  the  line  of  intersection  of  the  plane 
2x  —  3y  -\-  z  —  6  =  0  with  the  coordinate  planes. 

23.  The  equations  of  a  line  in  terms  of  its  direction 
cosines  and  the  coordinates  of  a  point  through  which  it 
passes.  —  Let  «,  /3,  and  7  be  the  direction  angles  of  the 


Fig.  13.  c 

line  and  Pt  a  point  through  which  it  passes.     Let  P  be 
any  point  on  the  line.     Then  from  the  figure 

x  —  xx  =  PXP  cos  a, 

y-yl  =  PXP  cos  ft 

z  —  z1  =  PXP  cos  7. 


238  ANALYTIC   GEOMETRY   OF  SPACE     [Ch.  IV,  §  24 

Solving  these  for  PXP,  and  equating  the  values,  we  have 
aj  -  asi     y  -  y\     z  -  Z\ 


cos  a         cos  0        COS  X 


PROBLEMS 


[22] 


1 .  What  form  will  these  equations  take  when  a  =  90°  ? 
when  a  =  90°,  and  £  =  90°  ? 

2.  Find  the  equations  of  a  line  through  the  point  (  —  1, 
2,  -3)  if 

(a)  a  =  60°,  /?  =  60°,      y  =  45°; 

(6)   a  =  120°,  (3  =  60°,     y  =  135°j 

(c)    cos  a  =  |  V3,  cos  /3  =  i,  cos  y  =  0. 

Show  that  the  given  values  are  possible  in  each  case  and  plot 
the  line. 

3.  Find  the  equations  of  a  line  through  the  origin,  equally 
inclined  to  the  axes. 

24.  Given  the  equations  of  a  line,  to  find  its  direction 
cosines.  —  The  method  is  best  shown  by  an  example.  Let 
the  equations  of  a  line,  reduced  to  their  simplest  form,  be 

x  =  5  z  —  6,  and  y  =  2  z  +  3, 


or 


Let 

cos  «        cos  p        cos  y 

be  the  equation  of  the  same  line.     These  equations  are  of 
the  same  form  and,  since  they  represent  the  same  line, 


X 

+  6 
5 

-V 

2 

3_ 

z 

-0 
1 

X 

-xx_ 

-V 

— 

Vi  _ 

z 

~zi 

x1  =  —  6,  yx  =  3,  and  z1  =  0, 


Ch.  IV,  §  24]  THE   STRAIGHT  LINE  239 

and  the  denominators,  5,  2,  and  1,  are  proportional  to  cos  a, 
cos  /3,  and  cos  7.  They  can  be  made  identical  with  them 
by  multiplying  by  a  suitable  factor  R. 

Then       cos  «  =  5  R,  cos  /3  =  2  R,  and  cos  y=R. 

1 


Then  by  [7]     25  R2  +  4  R?  +  R2  =  1,  and  i?  = 


Hence     cos  a  =  ,  cos  /3  =     t__ ,  cos  7  = 


30 


"30  V30'  V30' 

and  the  equation  can  be  written  in  the  form 

a?  +  6_y~3_g-0 
~5~"  ~~T~~~T~' 


V30       V30       V30 


PROBLEMS 


1.  Show  that,  if  the  equations  of  a  line  can  be  written  in 
the  form  x  =  mz  -\-  a,  and  y  =  nz  +  &>  they  may  be  changed 
into  the  form 

#  —  a  2/  —  &  2 


Vm2  +  w2  4- 1      Vm2  +  w2  + 1      Vm2  +  n2  +  1 

where    —       m        —  =  cos  a,   n  =  cos  /3, 

Vm2  +  w2  +  1  Vm2  +  ri2  +  1 

and  —  —  =  cos  y. 

Vm*  +  rr  +  1 

2.    What  form  will  the  equations   take,  if  their  simplest 
forms  are 

y  =  qx  +  c,  aj  =  e, 


240  ANALYTIC   GEOMETRY   OF   SPACE     [Ch.  IV,  §  25 

3.    Find  the  direction  cosines  of  the  lines  whose  equations  are 

(a)  2x  +  3y-2z-13  =  0, 
3x  +  6y-2z-24:  =  0. 

(b)  2x  +  2y-3z-    2  =  0, 
4ic—     y  —    z  —    6  =  0. 

(c)  2x  +  4y  +  3z  +    G  =  0, 

3x  +  6y  +  2z-    1  =  0. 

(d)  4y  +  3z  +    1=    0, 
3y-2z-12  =    0. 

25.  Equations  of  a  line  through  two  points.  —  Let  (xv 
yv  Zj)  and  (:r2,  yv  z2~)  be  the  two  points.  The  equation 
of  any  line  through  the  first  point  is  (by  [22]), 

x-xx  _  y-yx  _  g-  ^ 
cos  a        cos  /3        cos  7 

If  the  second  point  lies  on  this  line, 

xi  ~  x\      I/2  -  V\      Z2~  z\ 


cos  a  cos  /3         cos  7 

Dividing,  we  have,  as  the  equations  of  a  line  through  the 

two  points, 

as-asi  _y-yi_z-z\  1-23-1 


0C2  -vc\     2/2  -  2/1     22  -  Si 


PROBLEMS 


1.  Establish    equation   [23]   from    an    independent   figure 
without  using  equation  [22]. 

2.  Discuss  the  special  cases  of  [23],  when  x2  =  xx,  y2  =  yi, 
or  z2  =  Zj. 


Ch.  IV,  §  25]  THE  STRAIGHT   LINE  241 

3.  Find  the  equations  of  a  line  passing  through  the  points 

(a)  (0,  0,  -  2)  and  (3,  -  1,  0), 

(b)  (-  1,  3,  2)  and  (2,  -  2,  4), 

(c)  (2,  -  3,  1)  and  (2,  -  3,  -  1). 

4.  Find  the  equations  of  the  line  joining  the  origin  with 
the  intersection  of  the  planes 

3  x  -  2  y  +  z  +  4  =  0, 

a-  +  4?/  +  2z  =  0, 

y-3z-7  =  0. 

5.  Are  the  three  points  (1,  -  1,  2),  (2,  3,  -  1),  and  (3,  2,  2) 
in  a  straight  line  ? 

6.  Show  that  the  two  lines 

x-2  =  2  y-G  =  3z, 

and  ±x  -11  =  ±y  -13  =  3z 

meet  in  a  point,  and  that  the  equation  of  the  plane  in  which 
they  lie  is 

2x-6y  +  3z  +  U  =  0. 

7.  Show  that  the  line    4:X  =  3y  =  —  z   is  perpendicular  to 
the  line  3  x  =  —  y  =  —  4z. 

8.  Find  the  point  of  intersection  of  the  line 

2a -4  =  3?/  + 1  =  2  +  6 
with  the  plane  x  -{-  G  y  —  o  z  =  16. 

9.  What  is  the  equation  of  the  plane  determined  by  the 
point  (3,  2,  —  1)  and  the  line  2x  —  5  =  5y  +  l  =  z? 


CHAPTER   V 

QUADRIC   SURFACES 

26.  The  sphere.  — A  sphere  maybe  defined  as  the  locus 
of  a  point  whose  distance  from  a  fixed  point  is  constant. 

If  (#0,  y0,  Zq)  is  the  centre  and  r  the  radius,  the  equa- 
tion of  the  sphere  is  evidently 

(x  -  aco)2  +  (y  -  2/0)2  +  (s  -  zo)2  =  r2.  [24] 

If  the  centre  is  at  the  origin,  the  equation  becomes 

a?2  +  y2  +  z2  =  r2.  [25] 

Expanding  [24],  we  see  that  the  equation  of  every 
sphere  is  of  the  form 

oc2  +  y2  +  z2+  Gx  +  Hy  +  Iz  +  K  =  0,  [26] 

where 

a         _h        _i 

2 '  #o  _      2 '  z°  ~~      2' 


and  r  =  J  V£2  +  H2  +  I2  -  4JT. 

Every  equation  in  the  form  of  [26]  will  therefore  repre- 
sent a  sphere, 

real,  if   G2  +  H2  +  I2  -  4  K>  0, 

null,  if  G2  +  H2  +  I2  -  4  K=  0, 

imaginary,  if  G*  +  ff*  +  J*  -  4  ^T<  0. 

Comparing  [26]  with  the  general  equation  of  the  second 
degree, 

Ax2  +  By2  +  Cz2  +  Dyz  +  Ezx  +  Fxy+Gx  +  iry  +  Iz  +  K=0, 

242 


Ch.  V,  §  26]  QUADRIC   SURFACES  243 

we  see  that  the  general  equation  will  represent  a  sphere,  if 

D  =  U=F=Q,  and  A  =  B  =  0. 

A  sphere  may,  in  general,  be  passed  through  any  four 
points ;  for  the  substitution  of  their  coordinates  in  [26] 
will  give  four  equations  which  will,  in  general,  determine 
G,  J5T,  7,  and  if. 

PROBLEMS 

1.  Find  the  equation  of  a  sphere  with 

(a)  centre  at  (5,  —  2,  3),  radius  equal  to  1. 

(b)  centre  at  (2,  —  3,  —  6),  passing  through  the  origin. 

(c)  centre  on  the  Z-axis,  radius  a,  passing  through  the 
origin. 

2.  Find  the  centre  and  radius  of  each  of  the  following 
spheres,  when  real : 

(a)  ar  +  2/2  +  z2-2a  +  G?/-8z  +  22  =  0. 

(b)  x2  +  y2  +  z2  +  10x-4:y  +  2z  +  5  =  0. 

(c)  3a?  +  3tf  +  3z2+  12x  +  12 y  +  18 2  +  3  =  0. 

(d)  x2  4-2,2  +  32  +  60;=  0. 

(e)  ar°  +  2/2  +  z2  +  4.T  +  2/  +  5z  +  21  =  0. 

3.  Find  the  equation  of  the  sphere  passing  through  the  four 
points, 

(a)  (2,  5,  14),  (2,  10,  11),  (2,  5,  -  14),  (2,  -  10,  -  11), 

(6)  (0,  0,  0),  (2,  8,  0),  (5,  0,  15),  (-  3,  8, 1). 

4.  Find  the  equation  of  a  sphere  passing  through  the  origin 
and  concentric  with  the  sphere  through  the  points  (7,  7,  8), 
(_  i,  _  5,  -  8),   (-  5,  7,  -  6),  (3,  -  5,  10). 

5.  Find  the  equation  of  a  sphere  with  its  centre  at  the 
origin  and  touching  the  sphere 

&  +  f  +  z2  -  8  x  -  6y  +  2Az  +  48  =  0. . 


244  ANALYTIC   GEOMETRY  OF  SPACE        [Ch.  V,  §  27 

6.  Show  that  the  equation  of  the  sphere  whose  diameter  is 
the  line  joining  the  points  (a^  yx,  z^  and  (x2,  y2,  z2)  may  be  put 
in  the  form 

(x  -  x{)  (x  -  x2)  +  (y-  2/0  (y  -  y2)  +  (z  -  zY)  (z  -  z2)  =  0. 

7.  Show  that  the  equation  x2  +  y2  +  z2  =  r2  will  have  the 
same  form,  if  the  axes  are  turned  through  any  angle  without 
changing  the  origin. 

27.  Conicoids.  —  Any  surface  whose  equation  is  of  the 
second  degree  in  x,  y,  and  z  is  called  a  quadric  surface 
or  conicoid.  The  sphere  is  a  special  case  of  such  a 
surface. 

It  is  possible,  by  suitable  transformation  of  coordinates, 
to  reduce  the  general  equation  of  the  second  degree  in 
x,  y,  and  z  to  one  or  other  of  these  two  forms, 

(1)  Ax2  +  By2  +  Cz2  =  D, 

(2)  Ax2  +  By2  =  Cz, 

where  ^4,  B,  C,  and  D  may  be  any  quantities,  positive, 
negative,  or  zero.  But  for  our  present  discussion,  let 
neither  A,  B,  nor  0  vanish. 

The  locus  of  equation  (1)  is  evidently  symmetrical 
with  respect  to  each  of  the  coordinate  planes,  and  hence 
with  respect  to  the  origin.  Such  surfaces  are  therefore 
called  central  quadrics. 

If  D  =£  0,  equation  (1)  may  be  written  in  the  form 

±^±g±*?=l.  [27] 

a2     b2     c2  u     J 

If  D  =  0,  it  may  be  written 


Ch.  V,  §  28]  QUADRIC   SURFACES  245 

Non-central  quadrics  arc  included  under  equation  (2). 
It  may  be  written  in  the  form 

g±g=2~  [20] 

We   shall   now   investigate   the   forms  of   the   surfaces 
represented  by  these  equations. 

28.    The   ellipsoid.       ^  +  £?  +  ^=l._  The   surface  is 
a1      ¥      cl 

symmetrical  with  respect  to  each  of  the  coordinate  planes. 
Its  intercepts  on  the  X,  Y,  and  Z-axes  are  ±  a,  ±  b,  and 
±  c.     The  section  of  the  surface  made  by  the  X-Y- plane 

is  obtained  by  putting  2  =  0,  and  its  equation  is  -5+^  =  1, 

which  represents  an  ellipse  with  semi-axes  a  and  b.  The 
section  made  by  a  plane  parallel  to  this  coordinate  plane 
is  found  by  putting  z  =  zv     This  gives 


^  +  t„  =  l-Z4,  or  t + t =1 


a2      b2  c2  of  -j 


which    represents    an    ellipse,    in    the   plane  z  =  zv  with 

semi-axes  a^l \  and  byll  — -L,  the  centre  lying  on 

the  Z-axis. 

As  z1  increases  numerically  from  0  to  ±  c,  the  section 
diminishes  in  size,  until  when  zx  =  c  it  shrinks  to  a  null 
ellipse,  the  single  point  (0,  0,  <?).  As  zx  increases  nu- 
merically beyond  ±  c,  the  section  becomes  imaginary ; 
hence  the  surface  does  not  extend  beyond  the  planes 
z  =  o  and  z  =  —  c. 

Similarly,  the  sections  made  by  the  coordinate  planes 


246 


ANALYTIC   GEOMETRY   OF   SPACE        [Ch.  V,  §  28 


Y-Z  and  Z-X  and   by  planes  parallel  to  them   will   be 
found  to  be  ellipses  with  centres  along  the  X  and  Y- 


\ 

k 

\ 

\ 

\            \ 

f~ 

0 

/  !         1 

Y 

1/ 

LA 

\ 

/ 

'       \ 

'          v' 

Fig.  14. 


axes  respectively,  and  diminishing  in  size  as  the  cutting 
plane  moves  off  from  the  origin  to  the  points  where  the 
surface  cuts  the  axes. 


Fig.  li 


If  a  =  5,  the  equation  becomes 


f  +  zl 


1. 


a*      a*      c2 

Sections  made  by  planes  parallel  to  the  ^-I^-plane  will 
now  be  circles,  whose  centres  lie  along  the  Z-axis,  and 


Ch.  V,  §  29]  QUADRIC   SURFACES  217 

the  surface  is  an  ellipsoid  of  revolution  about  the  i£-axis. 

Similarly,   —  +  %  +  —  =  1    and    —  +  £-  +  —  =  1    repre- 
a2,      ¥      b1  a1      bz      a1 

sent   ellipsoids  of   revolution   about   the  X  and    y-axes 

respectively. 

If  a  =  b  =  c,  the  ellipsoid  becomes  the  sphere, 

£2  +  y1  +  22  =  #2« 

The  equation  ^-  +  ^--{-  —  =  —  1   is  not  satisfied  by  any 
a1      bl      c2, 

real  values  of  #,  ?/,  and  2  ;  it  may  be  said  to  represent  an 

imaginary  ellipsoid. 

29.    The  unparted  hyperboloid.    ^  +  |!  _??  =  1. —The 

a*      bl      e* 

intercepts  on  the  X  and  Y~-axes  are  ±  a  and  ±  5,  but  the 

surface  does  not  cut  the  Z-axis. 

The  section  of  the  surface  made  by  the  X-F-plane  is  the 

ellipse  —  +  *-  =  1 ;  the  section  made  by  the  r"-Z-plane  is 

(V*  22 

the   hyperbola    &-  — -  =  1,  with  its   transverse  axis,  2  5, 

along  the  Y"-axis  ;  the  section  made  by  the  Z-X-plane  is  the 

hyperbola  —  — -  =  1,  with  its  transverse  axis,  2  a,  along 
the  A-axis. 

The  sections  parallel  to  the  X-!F-plane  will  be  ellipses, 
with  their  centres  on  the  Z-axis ;  the  size  of  the  ellipses 
will  increase  without  limit  as  the  cutting  plane  recedes 
from  the  X-JT-plane  in  either  direction. 

We  have  now  sufficient  information  to  draw  the  figure. 

It  is  instructive,  however,  to  investigate  the  plane  sec- 
tions parallel  to  the  other  two  coordinate  planes. 


248 


ANALYTIC   GEOMETRY  OF   SPACE       [Ch.  V,  §  29 


The  section  made  by  the  plane  x  =  xv  parallel  to  the 
F-Z-plane,  may  be  written 


<r 


P  [1-=* 


C2     1 


xf\ 
aV 


=  1. 


If  xx  <  a,  this  represents  an  hyperbola  with  its  transverse 
axis  parallel  to  the  Praxis.     As  x1  increases  from  0  to  a,  the 


Fig.  16. 

semi-axes  both  approach  zero,  and  the  hyperbola  approaches 
a  pair  of  intersecting  lines.     When  x1  =  a,  the  section  is 

the  pair  of  straight  lines,  %-  —  z—  —  0,  in  the  plane  x  =  a, 

b2,      c2, 

intersecting  on  the  -X"-axis.  When  xx  >  a,  the  equation 
again  represents  an  hyperbola,  but  the  transverse  axis  is 
now  parallel  to  the  Z-axis.  As  xx  increases,  the  semi-axes 
increase  without  limit. 


Ch.  V,  §  29]  QUADRIC   SURFACES  249 

Similarly,  the  sections  made  by  planes  parallel  to  the 
Z-X-plane  will  be  found  to  be  hyperbolas,  the  transverse 
axis  being  parallel  to  the  X-axis,  when  the  distance  of  the 
cutting  plane  is  less  than  b,  and  parallel  to  the  Z-axis,  when 
the  cutting  plane  is  beyond  y  =  b  ;  the  transition  from  one 
set  of  hyperbolas  to  the  other  being  a  pair  of  intersecting 
lines  in  the  plane  y  =  b. 

wm^mmmmmmmmmmmmmmmmmmmmmmmmtmmm 


Fig.  17. 

The  surface  is  called  the  hyperboloid  of  one  sheet,  or  the 
unparted  hyperboloid,  extending  along  the  Z-axis.      The 

equations -+  \-  +  %  =  1  and  -r-  —  &-  +  \  —  1  represent 

a2      bl      <?  a2      ¥      cl 

unparted  hyperboloids,  extending  along  the  X  and  !F-axes 
respectively;  the  hyperboloid  in  each  case  extending 
along  the  axis  whose  coordinate  has  the  unique  sign  in 
the  equation. 

When    a  =  b,   the    equation   becomes    —  +  ^- =  1, 

H  a2      a2      c2 

which  is  the  equation  of  an  unparted  hyperboloid  of  revo- 


250 


ANALYTIC    GEOMETRY   OF   SPACE       [Ch.  V,  §  30 


lution  about  the  Z-axis.     The  equations «  +      H —  — 1 

^2        V2        z2  «2         *-        ^ 

and  —  —  *-  +  —  =  1  represent  hyperboloids  of  revolution 

^         ^        ^2 

about  the  X  and  Praxes  respectively. 

30.    The  biparted  hyperboloid.     --?£ --  =  !.--  The 

a2      oJ      e2 

intercepts  on  the  X-axis  are  ±  a,  hut  the  surface  does  not 
cut  the  other  axes. 


Fig.  18. 

The  section  of  the  surface  made  by  the  -X"-  F"-plane  is  an 
hyperbola,  with  its  transverse  axis  2  a  along  the  X-axis ; 
the  section  made  by  the  Z-X-plane  is  also  an  hyperbola, 
with  its  transverse  axis  2  a  along  the  X-axis ;  the  section 
by  the  ^F-Z-plane  is  imaginary. 

Sections  made  by  planes  parallel  to  the  Y"-Z-plane  are 
imaginary  for  values  of  x  between  +  a  and  —  a.  When 
x  =  ±  a,  the  sections  are  null  ellipses,  and  for  values  of  x 
numerically  greater  than  a  the  sections  are  ellipses,  in- 
creasing indefinitely  as  the  cutting  plane  recedes  from  the 
origin. 

The  sections  of  the  surface  made  by  planes  parallel  to 


Ch.  V,  §  30]  QUADRIC   SURFACES  251 

the  X-Y  and  Z-X-planes  are  hyperbolas,  and  it  may  also 
be  shown  by  the  aid  of  transformation  of  coordinates  that 
all  planes  through  the  X-axis  are  hyperbolas. 

This  surface  is  called  the  hyperboloid  of  two  sheets,  or 
the  biparted  hyperboloid,  extending  along  the  X-axis.    The 


Fig.  19. 

equations  __+|--=l  and   -__|  +  _  =  1  repre- 

sent  biparted  hyperboloids  extending  along  the    Y  and 
Z-axes  respectively. 

■■)■>£  yyw  y  L 

If  b  =  c,  the  equation  becomes  --  —  &-  — -  =  1,  which  is 

a1      c2      <? 

the  equation  of  a  biparted  hyperboloid  of  revolution  about 
the  X-axis.     The  equations, 

_E!_i_^_f!-l     and    _rr2_^4-22-l 
aJ      6^       <r  62       6^      c2 

represent  biparted  hyperboloids  of  revolution   about  the 
Y  and  Z-axes  respectively. 


252  ANALYTIC    GEOMETRY   OF   SPACE       [Ch.  V,  §  31 

31.    The    cone.      —  +  fr  — -  =  0.  —  A  cone,   or  conical 
az      bz      <r 

surface,  is  a  surface  generated  by  a  straight  line  passing 
through  a  fixed  point,  called  the  vertex,  and  always  touch- 
ing some  fixed  curve.  Any  position  of  the  generating 
line  is  called  an  element  of  the  cone. 

When  D  =  0,  we  have  seen  (Art.  27)  that  the  equation 

rp2l  qjA  iy& 

of  the  second  degree  reduces  to  —  ±  'f-  ±  —  =  0.     If  both 

aA      bz      cL 

the  positive  signs  are  used,  the  equation  is  satisfied  by 
the  coordinates  of  the  origin  only,  and  is  therefore  said 
to  represent  a  null  ellipsoid.  If  any  other  combination 
of  signs  is  used,  it  will  be  shown  to  represent  a  cone. 

rfiii  qjZ  yit 

Consider  the  equation    —  +  f- =  0. 

az      ol      <r 

The  section  made  by  the  X-I^-plane  is  a  null  ellipse; 
the  sections  made  by  the  Y-Z  and  Z-X-planes  are  pairs 
of  intersecting  lines. 

Sections  parallel  to  the  JT-I^plane  are  ellipses,  increas- 
ing indefinitely  in  size  as  the  cutting  plane  recedes  from 
the  origin.  Sections  parallel  to  the  other  coordinate 
planes  are  hyperbolas. 

Moreover,  if  (xv  yv  z^)  is  a  point  on  the  surface,  then 
any  other  point  (kxv  kyv  kz^)  on  the  line  joining  Px  with 
the  origin  will  also  lie  on  the  surface ;  hence  the  surface 
is  generated  by  a  straight  line  passing  through  the  origin, 
and  is  a  cone  extending  along  the  Z-axis. 

The  equations 

_^  +  g  +  j=0   and    tn-t  +  t     % 
a2      ¥      cl  a-      ¥      cz 

or  the  same  equations  with  their  signs  changed,  represent 
cones  extending  along  the  X  and  y-axes  respectively,  the 


Ch.  V,  §32]  QUADRIC   SURFACES  253 

cone  in  each  case  extending  along  the  axis  whose  coordi- 
nate has  the  unique  sign  in  the  equation. 

If  the  coefficients  of  the  two  terms  which  have  the 
same  sign  are  equal,  the  equation  will  represent  a  cone 
of  revolution  about  the  other  axis. 

32.  Asymptotic  cones.  —  The  equation  of  the  imparted 
hyperboloid  in  polar  coordinates  is 

2  /cos2  a      cos2 ft  _  cos2  y\  _  -j 


or 


V<*^ 


cos2  ft  _  cos2  y 


There  will,  therefore,  be  real  points  on  the  surface  for 
those  values  only  of  a,  ft,  and  y  which  make 

cos2«      cos2  6      cos2  y      r. 

Let  a',  ft' ,  and  yr  be  values  of  a,  ft,  and  7,  for  which 
this  expression  vanishes.  Then,  as  a,  ft,  and  7  approach 
a',  ft',  and  7',  the  value  of  p  will  increase  indefinitely, 
and  the  line  through  the  origin  whose  direction  angles 
are  a',  ft',  and  y'  may  be  said  to  meet  the  surface  at 
infinity.     Such  a  line  is  called  an  asymptotic  line. 

Since  the  equation  — — 1 - v— ?-  =  0   is   the 

a1  ¥  c2 

only  condition  which  must  be  satisfied  by  the  polar  coordi- 
nates of  the  points  on  all  the  asymptotic  lines,  it  must 
be  the  equation  of  the  asymptotic  cone,  which  contains 
all  these  asymptotic  lines  of   the  surface.      Multiplying 


254 


ANALYTIC   GEOMETRY   OF  SPACE       [Ch.  V,  §  33 


by  p'2   and   transforming  to   rectangular   coordinates,  it 
becomes 

*l + £.  _  ?! = n 

a2      62      c2 

Similarly  it  may  be  shown  that   the  asymptotic  cone 
of  the  biparted  hyperboloid  is 


x2      y1 


a2      b2      <? 


0. 


33.    The    paraboloids.       ^  ±  ^  =  2  cz.  —  The    surface 
—-  +  y~  —  2  cz  passes  through  the  origin,  but  does  not  cut 

(X  0 

the  axes  at  any  other  point.     Sections  made  by  planes 


Fig.  20. 


Fig.  21. 


parallel  to  the  X-F-plane  are  ellipses  whose  axes  increase 
as  the  section  recedes  from  the  origin.  Sections  made  by 
planes  parallel  to  the  other  coordinate  planes  are  parabo- 


Ch.  V,  §  33] 


QUADRIC   SURFACES 


255 


las,  which  have  their  axes  parallel  to  the  Z-axis.     This 
surface  is  shown  in  Fig.  20.    It  is  called  an  elliptic  parabo- 


2  2 

loid.     If  b  =  #,   --  -f-  &-  =  2  cz  represents  a  paraboloid  of 


a'      a 


revolution  about  the  Z-axis. 


Fig.  23. 


256  ANALYTIC   GEOMETRY   OF   SPACE       [Ch.  V,  §  33 


Let  the  student  discuss  the  form  of  the  surface  repre- 

-v>2  n \lil 

nted   by  the   equation  '—  —  ^-  =  2cz 
J  i  a2      b2 

hyperbolic  paraboloid.     (See  Fig.  22.) 


-v>2  n \lil 

sented   by  the  equation  '—  —  ^-  =  2cz.      It  is  called  an 
J  L  a2      b2 


PROBLEMS 

1.  Prove  that  in  both  the  elliptic  and  hyperbolic  parabo- 
loids the  sections  parallel  to  the  X-Z-plane  are  equal  parab- 
olas ;  also  that  the  sections  parallel  to  the  F-Z-plane  are  equal 
parabolas. 

2.  Show  from  the  results  of  problem  1  that  a  paraboloid 
may  be  generated  by  the  motion  of  a  parabola,  whose  vertex 
moves  along  a  parabola  lying  in  a  plane,  to  which  the  plane 
of  the  moving  parabola  is  perpendicular ;  the  axes  of  the  two 
parabolas  being  parallel,  and  (a)  in  the  elliptic  paraboloid,  their 
concavities  turned  in  the  same  direction ;  (b)  in  the  hyperbolic 
paraboloid,  their  concavities  turned  in  opposite  directions. 

3.  Show  that  an  ellipsoid  may  be  generated  by  the  motion 
of  a  variable  ellipse,  whose  plane  is  always  parallel  to  a  fixed 
plane,  and  which  changes  its  form  in  such  a  manner  that  the 
extremities  of  its  axes  lie  in  two  ellipses,  which  have  a  com- 
mon axis,  and  whose  planes  are  perpendicular  to  each  other 
and  to  the  plane  of  the  moving  ellipse. 

4.  Find  the  equation  of  the  cone,  whose  vertex  is  at  the 
centre  of  an  ellipsoid,  and  which  passes  through  all  the  points 
of  intersection  of  the  ellipsoid  and  a  given  plane. 

5.  Find  the  equation  of  the  cone,  whose  vertex  is  at  the 
centre  of  an  ellipsoid,  and  which  passes  through  all  the  points 
common  to  the  ellipsoid  and  a  concentric  sphere. 

6.  If  a,  b,  c  is  the  order  of  magnitude  of  the  semi-axes  of 
the  ellipsoid  in  problem  5,  and  if  the  radius  of  the  sphere 
is  b,  show  that  the  cone  breaks  up  into  a  pair  of  planes, 
whose  intersections  with  the  ellipsoid  are  circles. 


Ch.  V,  §  34]  QUADUIC   SURFACES  257 

34.  Ruled  surfaces.  —  A  surface,  through  every  point 
of  which  a  straight  line  may  be  drawn  so  as  to  lie  entirely 
in  the  surface,  is  called  a  ruled  surface.  Any  one  of  these 
lines  which  lie  on  the  surface  is  called  a  generating  line 
of  the  surface. 

The  cylinder  and  cone  are  familiar  examples  of  such 
surfaces.  We  shall  now  show  that  the  imparted  hyper- 
boloid  and  the  hyperbolic  paraboloid  are  also  ruled  surfaces. 

The  equation  of  the  unparted  hyperboloid  may  be 
written  in  the  form 

^_z2         _y* 
a*      c2  62' 


or 


e+3(H)=K)M> 

If  now  we  write  the  two  equations 

a      c      kx\        oj 

in  which  kx  may  have  any  value,  it  appears  that  every 
point,  whose  coordinates  simultaneously  satisfy  these 
equations,  will  satisfy  the  equation  of  the  hyperboloid, 
and  will  therefore  lie  on  the  surface.  But  these  two 
equations,  used  simultaneously,  are  the  equations  of  a 
line,  and,  from  what  we  have  shown,  that  line  must  lie 
wholly  in  the  surface.  But  since  hl  may  have  any  value, 
there  will  be  an  indefinite  number  of  such  lines,  and  it 
may  be  easily  shown  that  one  of  them  passes  through 
each  point  of  the  surface. 


258  ANALYTIC   GEOMETRY   OF  SPACE       [Ch  .V,  §  34 

In   the  same  manner  it  may  be  shown  that   there  is 
another  set  of  lines  whose  equations  are 


=K1+f) 


which  lie  wholly  in  the  surface.  A  line  of  this  set  may 
also  be  passed  through  any  point  of  the  surface.  Hence, 
through  any  point  on  this  ruled  surface,  there  may  be 
passed  two  lines  which  lie  wholly  in  the  surface.  Each 
line  of  one  set  cuts  every  line  of  the  other  set,  but  does 
not  cut  any  line  of  the  same  set. 

Let  the  student  show  that  the  hyperbolic  paraboloid  is 
also  a  ruled  surface.  Figures  17  and  23  show  the  two 
sets  of  generating  lines  on  both  these  surfaces.  None 
of  the  other  conicoids  are  ruled  surfaces. 

PROBLEMS 

1.  Prove  that,  if  a  plane  is  passed  through  a  generating 
line  of  a  conicoid,  it  will  also  cut  it  in  another  generating  line. 
Will  the  two  generating  lines  belong  to  the  same  set  ? 

2.  Prove  that  every  generating  line  of  the  ruled  paraboloid 

is  parallel  to  one  of  the  planes  -  ±  *-  =  0. 

a     b 

3.  Obtain  the  equations  of  the  generating  lines  which  pass 
through  the  point  (xlf  ylf  z^)  of  (a)  the  ruled  paraboloid, 
(6)  the  ruled  hyperboloid. 

4.  Prove  that  the  plane,  which  is  determined  by  the  centre 
and  any  generating  line  of  a  ruled  hyperboloid,  cuts  the  sur- 
face in  a  parallel  generating  line,  and  touches  the  asymptotic 
cone  in  an  element. 

5.  Show  that,  in  both  the  ruled  hyperboloid  and  the  ruled 
paraboloid,  the  projections  of  the  generating  lines  on  the  prin- 
cipal planes  are  tangent  to  the  principal  sections. 


Ch.  V,  §  35]  QUADRIC   SURFACES  259 

35.  Tangent  planes.  — A  tangent  line  to  a  surface  may 
be  defined  as  follows  :  Through  Px  and  P2,  two  adjacent 
points  on  the  surface,  draw  a  secant  line.  The  limiting 
position,  which  this  secant  approaches  as  P2  approaches 
Pv  is  called  a  tangent  line  to  the  surface  at  the  point  Pv 

Since  P2  may  approach  Px  along  the  surface  in  an 
indefinite  number  of  ways,  there  will  be,  in  general,  an 
indefinite  number  of  tangent  lines  at  any  point  of  a  sur- 
face. These  will,  in  general,  lie  in  a  plane  which  is  called 
the  tangent  plane  at  the  point  Pv 

We  shall  obtain  the  equation  of  the  tangent  plane  at 
the  point  Px  of  the  ellipsoid 

^-|_£?  +  ^  =  l. 

a2       b2      c2 

Transforming  this  equation  to  parallel  axes  with  the 
origin  at  Px  (by  [12]),  and  then  to  polar  coordinates 
(by  [6]),  we  have  as  the  equation  of  the  ellipsoid  in  polar 
coordinates  (origin  at  P2) 

2  /cos2  a      cos2/3      cos2  7 
9  \    a2  b2         ~J~ 

l2p(xicosa  ,  fficosff  (  giCQS7\=0> 
V      a2  b2  c2     J 

For  every  set  of  values  of  a,  /3,  7  in  this  equation  there 
will  correspond  two  values  of  p  ;  one  value  will  always  be 
zero,  which  agrees  with  the  fact  that  the  origin  is  a  point 
on  the  surface  :  the  other  value  is 


—  _  9. 


fxx  cos  a      yx  cos  ft      zl  cos  7 
V      a2  b2  c2 


.  cos2/3      cos2  7 
b2  c2 


260  ANALYTIC   GEOMETRY   OF   SPACE        [Ch.  V,  §  35 

which  gives  the  distance  from  Px  to  any  second  point  P 
of  the  ellipsoid,  measured  along  the  secant  whose  direc- 
tion angles  are  a,  /3,  7. 

Now  let  P2  approach  Px  along  the  ellipsoid ;  then  the 
secant  line  through  P1  and  P2  will  approach  as  its  limit- 
ing position  a  tangent  line  at  Pv  whose  direction  angles 
we  shall  call  «',  fi\  7'.     That  is,  as  P2  approaches  Pv 

,                          -.    x,  cos  a  ,  y,  cos  6  ,  z,  cos  7 
approaches    zero,    and   — l — h       ,«,       +  -1 — 5 — -    ap- 

Hence,   by   the 


theory  of  limits, 

xx  cos  (t!      yx  cos  ft      zx  cos  yr  _  ~ 
a*~  ¥~  c2        "    * 

If  (/o',  «',  ft,  7')  are  the  polar  coordinates  of  any  point 
on  any  one  of  the  tangent  lines  through  Pv  this  equation 
expresses  the  only  relation  which  must  hold  between  those 
coordinates,  and  is  therefore  the  polar  equation  (referred 
to  P1  as  origin)  of  the  locus  of  the  tangent  lines  through 
Pr     Multiplying  by  p'  and  transforming  to  rectangular 

coordinates  (by  [6])  we  have  ^  +  M  +  fb|  =  0.     Again 

a2        b2        c2 
transforming  to  the  original  origin  (by  [12]),  we  have 

a2        b2        c2  L      J 

as  the  required  equation  of  the  tangent  plane. 

Let  the  student  show  that  the  equations  of  the  tangent 
planes  to  the  hyperboloids, 

a2      b2      c2        ' 

are  «*g±tm-*&  =  1.  [31] 

a2        b2       c2  L     J 


Ch.  V,  §  3G]  QUADRIC  SURFACES  201 

the  paraboloids,  —  ±  ^-  =  2  cz, 

a1      ¥ 

are  *&±U&  =  e(.*  +  *d*  [32] 

36.  Normals. — The  line  perpendicular  to  the  tangent 
at  the  point  of  contact  is  called  the  normal  to  the  surface 
at  that  point. 

Its  equation  for  any  particular  surface  can  be  easily 
obtained  from  the  definition. 

PROBLEMS 

1.  Prove  that  every  tangent  plane  to  a  cone  passes  through 
the  vertex. 

2.  Prove  that  all  the  normal  lines  of  a  sphere  pass  through 
the  centre  of  the  sphere. 

3.  Show  that  the  length  of  a  tangent  to  a  sphere  from  the 
point  (#!,  yn  zx)  is  the  square  root  of  the  quantity  obtained  by 
substituting  (a^,  yx,  z{)  for  (xf  y,  z)  in  the  equation  of  the 
sphere. 

4.  Show  that  the  locus  of  points  from  which  equal  tangents 
may  be  drawn  to  a  sphere  is  a  plane.  This  plane  is  called  the 
radical  plane  of  the  two  spheres. 

5.  Prove  that  the  radical  planes  of  three  spheres  meet  in  a 
line.     This  line  is  called  the  radical  axis  of  the  three  spheres. 

6.  Prove  that  the  radical  plane  of  two  spheres  is  perpen- 
dicular to  their  line  of  centres. 

7.  Prove  that  the  radical  axis  of  three  spheres  is  perpen- 
dicular to  the  plane  of  their  centres. 

8.  Show  (from  its  definition)  that  the  tangent  plane  at  a 
point  Px  of  a  ruled  surface  contains  the  two  generating  lines 
of  the  surface  which  pass  through  Pv 


262  ANALYTIC   GEOMETRY   OF   SPACE       [Ch.  V,  §  37 

9.  Prove  that  every  plane  which  contains  a  generating  line 
of  a  ruled  surface  is  tangent  to  the  surface  at  some  point  on 
the  generating  line. 

37.  Diametral  planes.  —  The  locus  of  the  middle  points 
of  a  set  of  parallel  chords  of  a  quadric  surface  will  be 
found  to  be  a  plane.  This  plane  is  called  a  diametral 
plane. 

Let  av  fiv  yx  be  the  direction  angles  of  a  set  of  parallel 
chords  in  the  ellipsoid,  and  let  (V,  yf ',  z')  be  the  coordi- 
nates of  the  middle  point  P'  of  any  one  of  these  chords. 

Transform  the  equation  of  the  ellipsoid  to  polar  coor- 
dinates with  P'  as  origin.  Its  equation  (by  [12]  and 
[6])  is 

2 /cos2  ft      cos2/3      cos27\      9    /Vcos  a      ?/'cos/3      g'cos  y 
P\~aT+     b*     +     c2    J        P\     a2  V  <?     , 

~'2        f/2        ~'2 

^  a2  +  62  "*"  c2 

The  two  values  of  p  given  by  this  equation  are  the  two 
distances  from  the  origin  to  the  ellipsoid,  measured  along 
a  line  whose  direction  angles  are  «,  /3,  y.  If  «,  /3,  7  have 
the  particular  values  av  /3V  yv  the  two  distances  are 
equal  but  opposite  in  sign  ;  the  sum  of  the  roots  of  the 
equation,  regarded  as  a  quadratic  in  p,  will  be  zero,  and 
(by  Introduction,  Art.  8), 

x'  cos  «1      y'cosfil      g'cosy1  _  « 
a2  ft2  c2       "    " 

But  #\  y,  2'  are  the  coordinates  of  any  point  on  the 
required  locus,  referred  to  the  original  axes.     Hence 

00  COS  ai    ,   V  COS  pi    .   Z  COS  Yi  _  q  roo-l 

a2  62  c2  L     J 


Ch.  V,  §  37]  QUADRIC   SURFACES  263 

is  the  equation  of  the  diametral  plane  bisecting  the  chords 
of  the  ellipsoid  whose  direction  angles  are  av  /3V  7r 

The  locus  is  evidently  a  plane  passing  through  the 
centre  of  the  ellipsoid,  and  it  is  easily  seen  that  any 
plane  passing  through  the  centre  will  be  a  diametral 
plane  bisecting  some  system  of  parallel  chords. 

Let  the  student  show  that  the  equations  of  the  diametral 
planes  of  the  hyperboloids, 

the  paraboloids, 

a2      b2  <*2  &  L     J 

From  the  last  equation  it  appears  that  the  diametral 
plane  of  a  paraboloid  is  always  imrallel  to  the  axis  of 
the  paraboloid. 

The  line  of  intersection  of  any  two  diametral  planes  is 
called  a  diameter.  All  diameters  of  the  central  quadrics 
evidently  pass  through  the  centre  ;  in  the  paraboloids  they 
are  parallel  to  the  axis. 

It  may  be  shown  that  in  the  central  quadrics  there  are 
three  diameters  which  are  so  related  that  the  plane  of  any 
two  bisects  all  chords  parallel  to  the  third.  Such  diame- 
ters are  said  to  be  conjugate  to  each  other  ;  and  the  plane 
through  any  two  of  them  is  conjugate  to  the  third. 

PROBLEMS 

1.  Obtain  the  equation  of  the  diametral  plane  conjugate  to 
a  diameter  through  the  point  (xXi  yly  z^  of  (a)  the  ellipsoid, 
(b)  the  hyperboloids,  (c)  the  paraboloids.     (See  [6].) 


264  ANALYTIC    GEOMETRY   OF   SPACE       [Ch.  V,  §  38 

2.  Show  that  the  tangent  planes  at  the  extremities  of  a 
diameter  are  parallel  to  the  diametral  plane  conjugate  to  the 
given  diameter. 

3.  Show  that  the  relation  which  exists  between  the  direction 
cosines  of  any  pair  of  conjugate  diameters  is 

cos  ax  cos  a2      cos  fa  cos  fa      cos  yj  cos  y2  _  n 
cr  b~  c 

4.  Prove  that  the  diametral  plane  of  a  sphere  is  perpendic- 
ular to  the  chords  which  it  bisects,  and  that  conjugate  diame- 
ters are  perpendicular  to  each  other. 

5.  Prove  that  every  plane  which  passes  through  the  centre 
of  a  central  conic,  or  is  parallel  to  the  axis  of  a  non-central 
conic,  is  a  diametral  plane,  and  find  the  direction  cosines  of  the 
chords  which  it  bisects. 

38.  Polar  Plane.  —  The  locus  of  points  which  divide 
harmonically  secants  drawn  from  a  fixed  point  to  a  quadric 
surface  will  be  found  to  be  a  plane.  It  is  called  the  polar 
plane  of  the  given  point  with  respect  to  the  quadric  sur- 
face.    The  fixed  point  is  called  the  pole  of  the  plane. 

We  shall  obtain  the  equation  of  the  polar  plane  of  the 
point  Px  with  respect  to  the  ellipsoid 

x2      y2      z2       . 
a2      b2      c2 

We  have  seen  that  the  polar  coordinate  equation  of  the 
ellipsoid,  referred  to  Px  as  origin,  is 

i 

YC0S2«        COS2/3        COS27\  /^COStf        y.COS/3        Z.COSy 

x_l      iL2      z2 

a2  ^  62       ca      U 


Ch.  V,  §  38]  QUADRIC   SURFACES  265 

Through  the  origin  Px  pass  a  secant  whose  direction 
angles  are  «',  ft',  y'.  Let  the  points  where  this  secant  cuts 
the  ellipsoid  be  P2  (p2,  a',  /3',  7')  and  Ps  (j>3,  «',  /3',  7'), 
and  on  it  locate  a  point  P'  (//,  a',  /?',  7')  such  that 

p  pi  _  2P\P2  X  P\P*     or  ,/  _  1m.. 
P1P*  +  P1PZ'  P         R2  +  ps 

Then  /?3  and  /93  are  evidently  the  roots  of  the  equation 

0/cos2«'      cos2  8'      cos27f\ 


Hence  (by  Introduction,  Art.  8) 


a2  "*"  62  "*"  c2 
P  '=- 


xx  cos  a'      yx  cos  /3'     2; x  cos  7' 


-  + 


This  is  an  equation  connecting  the  polar  coordinates  of 
P\  and  is,  therefore,  the  polar  equation  of  the  desired 
locus.  Transforming  to  rectangular  coordinates  and  to 
the  original  origin,  Ave  have,  as  the  equation  of  the  polar 
plane, 

aZ     +     bl     +    C2     ~  1-  Ldt)  J 

Let  the  student  obtain  the  equation  of  the  polar  plane 
of  a  point  with  respect  to  each  of  the  quadric  surfaces. 


266  ANALYTIC   GEOMETRY   OF   ST  ACE        [Ch.  V,  §  38 

PROBLEMS 

1.  Prove  that  the  polar  plane  of  1\  with  respect  to  any 
quadric  surface  passes  through  the  points  of  contact  of  all  the 
tangent  lines  from  Px  to  the  surface. 

2.  Prove  that  the  polar  planes  of  all  points  in  a  given  plane 
pass  through  the  pole  of  that  plane ;  and,  conversely,  the  poles 
of  all  planes  passing  through  a  given  point  lie  on  the  polar 
plane  of  that  point. 

3.  Prove  that  the  polar  planes  of  all  points  on  a  given 
diameter  of  a  quadric  surface  are  parallel  to  the  tangent  plane 
at  the  extremity  of  the  diameter. 

4.  Prove  that  the  polar  plane  of  Px  with  respect  to  a  sphere 
is  perpendicular  to  the  diameter  through  Pv 

5.  Prove  that  in  the  sphere  the  product  of  the  distance  of 
the  pole  from  the  centre  and  the  distance  of  the  polar  plane 
from  the  centre  is  equal  to  the  square  of  the  radius. 

6.  Prove  that  the  distances  of  two  points  from  the  centre  of 
a  sphere  are  proportional  to  the  distances  of  each  from  the 
polar  plane  of  the  other. 

LOCI    PROBLEMS 

1.  Find  the  locus  of  points  which  are  equally  distant  from 
two  intersecting  planes.  Show  that  it  consists  of  two  planes 
which  are  perpendicular  to  each  other. 

2.  Show  that  the  locus  of  a  point,  the  sum  of  the  squares  of 
whose  distances  from  any  number  of  points  is  constant,  is  a 
sphere. 

3.  A  point  moves  so  that  the  sum  of  the  squares  of  its  dis- 
tances from  the  six  faces  of  a  cube  is  constant;  show  that  its 
locus  is  a  sphere, 


Ch.  V,  §  38]  QUADRIC    SURFACES  267 

4.  ^l  and  B  are  two  fixed  points,  and  P  moves  so  that 
PA  =  nPB-,  show  that  the  locus  of  P  is  a  sphere.  Show  also 
that  all  such  spheres,  for  different  values  of  n,  have  a  common 
radical  axis. 

5.  Show  that  the  locus  of  the  point  of  intersection  of  three 
mutually  perpendicular  tangent  planes  to  an  ellipsoid  is  a  sphere 
about  the  centre  of  the  ellipsoid,  whose  radius  is  Va2  +  b2  +  c2. 

6.  Show  that  the  locus  of  the  point  of  intersection  of  three 
mutually  perpendicular  tangent  planes  to  a  paraboloid  is  a 
plane. 

7.  Find  the  locus  of  a  point  whose  distance  from  a  given 
point  bears  a  constant  ratio  to  its  distance  from  a  fixed  plane. 

8.  Three  fixed  points  on  a  line  lie,  one  in  each  coordinate 
plane ;  find  the  locus  of  any  fourth  fixed  point  of  the  line. 

9.  Show  that  the  locus  of  the  points,  which  divide  in  any 
g.ven  ratio  all  straight  lines  terminated  by  two  fixed  straight 
lines,  is  a  plane. 

10.  A  line  of  constant  length  has  its  extremities  on  two 
fixed  straight  lines  j  show  that  the  locus  of  its  middle  point 
is  an  ellipse. 


ANSWERS, 


Page  10 


*■  «  (!•  §)•  (-!•  I)-  (-!•  -!)•  (I-  -|> 

|V2,o).   (o,  |V5).  (~|V2,  o)»  (o,  -|V2~)' 


(ft) 

5.  (a)   (a,  0),  (6,  c),   (a  +  b,c). 
(b)   (a,  0),  (0,  c),  (a,  c). 

6.  (a)    (0,0),  (0,0),    (|,   ^ay 

w  (1-  o).  (-1-  o),  (o,  *.). 

(0    (f  V3,   0),    (-gvj.   |),   (~§A    -I)- 

Page  14 

2.    \'34>    Vl30,   2\/29.  3.    5V3,    2Vl3,    V7. 

4.  Va2  +  62,    Va2  +  62  +  a&V2. 

Page  18 

2.  (-19,  -16).  12.    f,  --V-. 

3.  (-11,2).  13.    (-9,6). 

10.   (11,5).  14.    (a)    (V,  -|).  (6)    (1,  -1). 

Page  23 

1.  6x-4?/  + 19  =  0.  6.    2  x2  +  2  ?/2  +  14  x  -  19  y  +55  =  0. 

2.  y  =  3x.  7.    x  +  ?/-  10  =  0. 

3.  x2  +  ?/2  +  6  x  -  8  y  =  0.  8.    x2  -  3  ?/2  =  0. 

4.  24x2  +  25 ?/2  -  250 x  + 625  =  0.     9.    8x-2?/+17=0. 

5.  x2  +  ?/2  -  5  a:  +  5  y  +  5  =  0.       10.    x2  +  y2  -  x  -  y  =  0. 

Page  32 
3.    2V5;   |V2;   jVl70,  6.    &<£;   6>i;   &  =  fr 

5.   6,  7.   x2  -  i/2  =  0. 

269 


270  ANSWERS 

Page  36 

2.  (a)  5x  +  8y  =  7.  (c)   x  -  4  =  0. 
(6)  3x-4y  =  0.  (tf)y-5  =  0. 

3.  x~Sy  =  S.  4.   Yes.     No. 

5.  2/i  (Xa  -  £3)  +2/2 (£3  -  »i)+  2/3  (xi  -  x2)  =  0. 

6.  39  x  -  79  y  =  200. 

7.  Equations  of  medians,   x  —       y  —  1  =  0, 

x  +    2  y  +  1  =  0, 
x  -  13  y  -  9  =  0. 
Point  of  intersection,      (i,  —  f). 

8.  Equations  of  diagonals,  bx  +  ay  =  ab, 

bx  —  ay  —  0. 

Point  of  intersection,      [-»    -  V 
V2     2J 

Page  40 

1.  \/3x--j/=-6(V3  +  l). 

2.  (a)    >/3x-3y  =  -  18.  (6)   3x  -  5  y  =  25.        (c)  x-y=-3 

3.  7V3x  +  7y  =  llV3-2.  4.   -2. 

Page  43 
2.    (a)  a  =  10,  b  =  -  f ,  I  =  J.  4.    (a)      x  -  8  y  +  5  =  0. 

(6)  «  =  -*,  b  =  l,l  =  l  (6)2*-     y-2  =  0. 

(c)  a  =  0,  6  =  0,  Z  =  -  4.  (c)    3  x  +  5  y  =  0. 

(d)  a  =  —  4,  &  =  00,  Z  =  ao 

Page  46 

2.    tan-1(2^|).  3.    135°,  tan-!(7),  tan-i(6$),  tan-i(lf). 

4.  Exterior  angle  between  first  two  lines,  tan_1(—  |f).     Opposite  inte- 

rior angles,  tan"1  (||),  tan-!(- -2^). 

Page  47 

2.  7x-3y  =  ll. 

3.  (a)  y  =  0,  4  x  +  y  -  24  =  0,  2  x  -  y  =  0. 
(&)  x  -  4  y  =  0,  x  +  2y-6  =  0,   x  -  4  =  0. 

(c)  x  -  3  =  0,  x  -  4  y  +  11  =  0,  x  +  2  y  -  10  =  0. 

(d)  4  x  -  5  y  =  0,  x  +  y  -  6  =  0,  8  x  -  y  -  24  =  0. 

Page  49 

\.    (6  +  5V'3)x-.-3y  =  0.  g.    7  x  -  3  y  +  5  =  0. 


ANSWERS 


271 


Page  51 
1.    (a)  x  +  y/3  y  -  10  =  0.  (6)  x  -  V3  ?/  +  10  =  0. 

(c)  VSx-y  +  10  =  0.  (d)  a  +  y  =  0. 

0)   4  x  +  (2  +  2  V3)y  -  4  V2  =  0. 
(/)  2  x  -  (  V2  -f  V'0)y  -  24  =  0. 


1.    T33\/l3. 


4.    4z  +  3?/-3  =  0 


Page  54 
2.    -  1. 

(H.  -tt);  7|. 


3.    11$. 

6.    T2T53  \/lT3  ;  the  first. 


Page  57 
1.   x  +  7y  +  3  =  0',7x—y 


17=0. 


2.  llz-35y=0. 

3.  49  x +  98?/- 272  : 

4.  3  x  +  2  ?/  +  7  =  0. 


Page  58 

0  ;  15  x  -  18  y  -  320  = 
5.    V3z-i/  +  3-V3 


0  ;  4  z  +  5  ?/ 

:0.        6.     10  X- 


3=0. 
3  y  +  4  =  0. 


Page  60 


1.    32. 


3.    3263 


Page  60.    General  Problems 


2(  6        41"\  .     /"78      23\ 
•     VT7>  T7J>    UT»   IT) 


3. 
4. 
6. 
7. 
9. 

11. 


15. 


2  x  +  5  y  =  40 ;  18  a  +  5  y  =  120  ;  6(2  +  V7>  +  5(2  ±  y/l)y  =  120. 

5x-y+10  =  0.  5.    2x  +  3?/  +  12  =  0;6x  +  ?/-12  =  0. 

(a)  x  —  4  y  =  0. 

(31,  5±|V3).  8.    (10,51). 

(16f,  -0&);  (4|,   -jf).         10.    (8A*  1H)J  4tt,  3&). 

/5VI97  ±  3V82     4V197  ±  14  V82\ 


V  Vl97 +V82 
/4V29  +  3V82t 
*    V29+V82 
/4V29  +  5V197 

V  V29  +  VI97 

5  x  +  y  =  26. 


Vl97±\/82  / 
7V29  +  7V82\. 
V29  ±  V82  /  ' 
-  7\Z29±2Vl97 
V29+VI97 

(Cl  -  C2)2 

2  (mi  —  m2) 


17 


18.    5. 


21.    (5,5). 


Page  70 
2.    4  x  -  3  y  +  15  =  0. 


272 

ANSWERS 

Page 

71 

1. 

Sx+  Ty  -  10  = 

:0. 

5.    (- 

-4,  2). 

3. 

xy  =  5- 

6.    11 

tan-i(- 

■2). 

4. 

tan-1  f. 

Page 

74 

2.    (a) 
(&) 

p2: 

«2&2 

4 

.    (a)  x2  +  y2-  ay 

(b)  (x-b)Vx* 

(c)  x2+y2-ax- 

=  0. 
+  *■  = 

ax. 

a2  cos2  tf  +  &2  sin2  0 
2  m  cos  0 
sin20 

-  a  Vx2 

+  y~ 

=  0. 

(«) 

P2-- 

a2 

(d)  (x*  +  y*y  = 

(e)  x2  +  y2  +  bx- 

a2(x2  - 
-aVs2 

■  y2)- 

+  2/2 

cos  2  0 

=  0. 

w 

(/) 

(so 

P2-- 
P     = 
P     = 

P     = 

=  a2  cos  2  0. 
=  a  sin  2  0. 
=  —  a  cos  0. 
2  a  -  cot2  0 

COS0 

(/)  (*2  +  ?2)8  = 
fa)  x2  -  ?/2  =  a2. 

(0    (x2  + y2)3  = 

4  a2xV 

2(x2  +  2 
4a¥. 

2 

xy  — 

*/2)2. 

(/O 

P     = 

=  a(l  —  2cos0). 

( j)  z3  +  ^2  -  2 

a*/2  =  0 

Page  78 

1.  (a)  x2  +  2/2  +  4  x  -  6  */  -  23  =  0. 

(b)  x2  +  2/2  +  6x  +  8y  =  o. 

(c)  x2  +  */2  -  10  x  -  6  ?/  +  33J-f  =  0. 

(d)  x2  +  y2  -  36  x  -  32 -y  +  480  =  0,  or 

x2  +  i/2  -  4  x  -  8  y  -  80  =  0. 

(e)  19  x2  +  19  y1  +  2  x  -  47  */  -  312  =  0. 
(/)  3  x2  +  3  y*  -  13  x  -  11  y  +  20  =  0. 
(flO  x2+«/2-x-42/-6  =  0. 

(h)  3x2  +  3y2-  114  x-  64  y  +  276  =  0. 

(*)  x2  +  2/2_(5±9VT|)x-(3±5VH)2/-18±30VH: 

2.  (a)   (-4,  3);    V35.  (c)   (0,    -3);    5. 
(6)  Imaginary.                        (d)   (£,  0);  |Vl45. 

6.    6  x  +  3  y  -  10  =  0. 

Page  82 

1.  (a)  3  x  +  4  y  =  25  ;   4  x  -  3  y  =  0. 
(6)  Indeterminate. 

(c)  3  x  +  7  y  =    93  ;  7  x  -  3  y  =  43. 
3  x  -  7  */  =    65;  7  x  +  3  */  =  55. 

(d)  6  x  +  5  */  =  114  ;  5  x  -  6  y  =  -  27. 
6  x  -  5  y  =    44;  5  x  +  6  ?/  =  57. 

2.  45°.  5.    V4L 


ANSWERS  273 

Page  84 

1.  (a)  (21±4V51)x+(28^3\/51)y  =  350. 

(b)  x  +  5  y  -  28  =  0  j  5  x  -  y  +  10  =  0. 

(c)  2  x  -  //  =  15  ;   58  x  +  71  */  =  335. 

2.  (a)  6  x  +  8  ?/  -  49  =  0.  (c)  14  x  +  3  y  =  55. 
(6)  2  x  -  3  ?/  +  9  =  0. 

3.  xi    x  +  yi    y  =  r2. 

Page  85 

1.  (a)  3a:-2y±7Vl3  =  0.  (6)  2x  +  3  y  ±  7V13  =  0. 

2.  3x  +  ?/  +  9±3VT0  =  0.  3.    1  +  1  =  1. 

r-  a2      b~      r1 

4.  A:  =  3G±20v/G. 

5.  A2E2  +  B2D2  +  4BCE-2ABDE-4A2F-4  C2  +  4  ACD-4  B2F=0. 

Page  86 

1.    f>/2T.  2.    x2  +  ?/2-f  52  x  -  21  y- 265  =  0. 

3.  2  x2  +  2  ?/2  -  13  x  -  6  ?/  +  15  =  0. 

64V26  +  375\2  ,  /         2  V26 -f  435  \2_  /  301 


no/      V: 


25V26+170'        \        25V26  +  170/        \25>/26  +  170/ 

6.    -^-.  7.    2x  +  y  =  2;  x  -  2  y  =  0  ;    (0,0),   (f,  |);   tan"*  2. 

2x01 

Page  89 

3.  Perpendicular  bisector  of  the  line  joining  the  two  points. 

5.  Circle  of  radius  r  about  (xi,  y\). 

6.  Perpendicular  bisector  of  the  line  joining  the  two  points. 

7.  Bisector  of  the  angle  between  the  lines. 

8.  Circle  about  the  centre  of  the  square. 

9.  Circle  whose  centre  is  on  the  line  through  the  fixed  point,  perpendic- 

ular to  the  fixed  line. 

11.  Circle  whose  centre  is  on  the  base  of  the  triangle,  extended. 

12.  Circle  whose  centre  is  at  the  centre  of  the  triangle. 

14.  Circle  whose  centre  is  at  the  intersection  of  the  two  lines. 

15.  Circle  whose  centre  is  on  the  line  OX. 

18.  Line  through  the  centre  of  the  base  and  the  centre  of  the  altitude  of 

the  triangle. 

19.  A  straight  line. 

20.  Two  lines  through  the  origin. 

21.  A  line  through  the  origin. 

22.  A  circle. 

23.  x2  +  y2  -  rVx2  +  y2  =  ry. 


274  ANSWERS 

24.  A  diagonal  of  the  rectangle. 

25.  A  diagonal  of  the  parallelogram. 

30.  x2  +  y-2  _  _1«_  {XlX  +  yiy)  =  ( tULz 

m  +  n  \m  +  n, 

31.  xix  +  yxy  =  r2. 

32.  An  equal  circle  tangent  to  the  given  circle  at  the  fixed  point. 

33.  (x12+yl2-r2)[(x-xO2+(y-yl)2]  +  2k2(xlx  +  y1y-x12-y{2)  +  tf  =  0. 

34.  A  line  parallel  to  the  fixed  line. 
35     A  circle. 

Page  103 

1.  (a)  y2  =  -  2  mx.  (6)  x2  =  2  my.  (c)  x2  =  —  2  my. 

2.  y2  =  2  mx  +  m2.  4.    4  x2  =  —  9  y. 

3.  G/-/3)2  =  2m(x-a).  5.    x  =  -2  ;  (2,  0);  8. 

Page  110 

1.  (a)  a  =  3,  &  =  2,  c  =  V5,  e  =  |V5,  x  =  ±  — -• 

V5 

(6)  a  =  3,  6  =  2,  c=V5,  e  =  | V5,  y=±-iL 

v5 

(c)  a  =  |V30,  6  =  1V5,  c  =  |  ^3,  e  =  \ VTO,  *  =  ±  f  V3. 

2.  (a)  4  *2  +  9  y2  =  36.  (d)  16  x2  +  25  y2  =  400. 
(6)  3  x2  +  4  ?/2  =  36.  (e)  16  x2  +  25  i/2  =  400. 
(c)  5  x2  +  9  ?/2  =  180.                          (/)    8as*  +    9y2  =  U52. 

5.    3x2  +  71/2  =  55.  6.    il^)l+iyj=jy  =  l. 

a2  b2 

Page  115 

1.  (a)  a  =  5,  6  =  1,  c  =  V26,  e  =  |  V20,  &  =  ±  — ,  a;  ±  5  y  =  0. 

V26 

(6)  a  =  2,  6  =  3,  c=Vl3,  e  =  |Vl3,  ac=±-4=,  3  x  ±  2  y  =  0. 

Vl3 
(c)  a=Vl0,  6=2,  c=VH,  e  =  iV35,  x=±fVl4,  2x±VlO>=0. 

2.  («)  4  x2  -  9  i/2  =  36.  (d)  16  x2  -  9  y2  =  144. 
(6)  3  x2  -  y2  =  9.  O)  9  x2  -  16  y2  =  144. 
(c)  5  x2  -  4  y2  =  125.                          (/)  72  x2  -    9  y2  =  800. 

3     (s-«)a_(y-fl)8  =  ;L  4    Impossible. 

a2  62 

Page  118 

1.  4  x2  —  y2  =  -  4  ;    a  =  1,   6=2,   e  =  |V5;    latus  rectum  =  1 ;    foci, 

(0,  ±  V5) ;  directrices,  y  =  ±  %  V5. 

2.  V2.  4.   2xy  =  a2. 


ANSWERS  275 

Page  123 
1.    -y-v^;   f^-  2-    |Vl3±|V66. 

Page  130 

1.    (a)  3  ac  +  8  y  =  19 ;  8x-3y  =  2. 

(b)   3x  +  y  =  7  ;   x  -  3  y  =  9. 
fr)    x+2y  =  -6;   2»-y  =  18. 
(d)   5  x  -  0  y  =  -  8  ;   6  &  +  5  y  =  27. 

2-    (a)  ¥;    -I-         (&)    -l!   6.  (c)   -12;  3. 

8.    (a)  y  =  4;  3&  +  2y  =  17.  4.    (a)   12  x  +  25  y  =  100 

(b)  x-y=-l;   x  +  Sy=-9.  (6)x4^3. 
(c)»  +  3y  =  5;as-3y  =  -7.  (c)  y  =  3. 

5.    (a)  |\/73;   ^V73.         (6)  fVTO;   2vT6. 

(c)  6 a/5;  3V5.  6  tan-i(±  3). 

Page  132 

1.  x-22/iVTT^O.  5.    (™,    ±ml; 

2.  18  x  +  27  y  =  88.  V2  ' 

3.  5x+2/_ V      Va2+&2  vW&2/ 

4.  /3  =  ±  V&2  -  a2Z2.  (x        «2       ,    ±        ^       \ 

V      Va2  -  &2  Va2  -  bV 

Impossible  in  hyperbola  when  b  >  a. 

V      V2  V2/ 

8.  5  y  ±  xVlE  ±  4\/l0  =  0.     Four  tangents. 

9.  1  y  ±  2  x  V35  ±  4V91  =  0.     Four  tangents. 

10.   x  ±  yy/S  +  6  =  0.     Two  tangents..  12.    (a)  4.         (&)  3. 

Page  148 

1.  3  x  -  8  y  =  0  ;  x  -  3  y  =  0.     5.    20  x  +  33  y  =  125. 

2.  */  +  9  =  0.  6.   8x  +  45y  =  0;(      45_,    § — V 

3.  2*V§  +  3y  =  0.  7.   ^V3.  ^±Vl51      TVl61/ 

4.  x  +  2  y  =  8.  8x-y  =  l. 

9.  (±  i Vl5,   T  i\/l5);   (±  §  Vl5,   T  t^ Vl5).  10.    h=-  12- 

Page  158 
1.    (a)  x  -  8  y  =  16.  (c)  15  x  +  16  y  =  -  24. 

(b)  x  +  2  g  =  -  6.  Cd)  x  +  5  =  0. 


276 

6.  (-If,  H). 


ANSWERS 

4.    (-10,4).  5.    (-^,   W\. 

7     /       a2b2Xi  a*b2y\       \ 

\62a;i2  +  a-y{2  b'2x{2  +  cPyi2) 


1.  Imaginary  ellipse 

2.  Real  ellipse. 

3.  Two  intersecting  lines. 

4.  Hyperbola. 

5.  Two  parallel  lines. 


Page  182 

6.  Parabola. 

7.  Two  coincident  lines. 

8.  Point. 

9.  Point. 

10.    Hyperbola. 


Page  186 
4.    x-. 


2k 


2.   The  directrix. 

5.    (a)  x2  +  y2  =  a2;  (6)  x2  +  y2  =  a2  ;  (c)  x  =  0. 

7.    The  asymptotes. 


v2 

'  6* 


e.  ^+r-2. 


10.    ft2*2  +  a-Y2  =  62c2. 
12.    2/^-— x. 


a4      6*     a2     62 
11.    («)  An  ellipse  ;    (6)  A  parabola. 

14.    25  x2  +  16  y2  -  48  y  -  64  =  0. 


15.    2  r  VxJ  +  yl  —  2  xxx  -  2  ?/ii/  +  Xi2  +  y{2  —  r2  =  0. 


16.    x2 


x  +  fy-|  =  0. 


3  x2  —  ?/2  —  2  ex  =  0.     (Take  the  origin  at  the  vertex  of  the  smaller 
angle. ) 


22.  A  parabola. 

24.  2  xy  —  y\X  —  Xiy  =  0. 

27.  y2  =  -2  mx. 

30.  x=-r>L±m. 

2 

32.  (x2  +  y2)2  =  d2x2  +  b2y2. 

34.  4  &!2xV2  -  4  rtiV  =  «i2&i4- 

36.  A  directrix. 


23.  x2  +  ?/2=-. 

26.  (x2  +  y2-2  ax)2  =  a2(a-x)2+a'V2 

28.  ab'2Vx2  +  2/2  =  62x2  +  d2y2. 

31.  x2  +  y2-(c2  +  m'2)x  +  c2  =  0. 

a4  64 

35.  Circle  with  radius  a  +  b. 

37.  *f^+W1 +  *  +  *  =  !■ 


38.    *-  +  ^=l. 
a2     a4. 

ft2 


ANSWERS 


Page  211 

V^O;   (6,    -10,  20).       2.    (*L±**±**     Vi +  **  +  &,   *i-r*2  +  z*\ 

\  3  3  3  / 


COS" 

ai 

V3. 

Page 

214 

COS 

■/s 

_     3 

Vn 

_  -4 

,    cos  7 

1 

VTT 

7 

a  =  P  =  7  = 

cos  «  = =, 

Vll 


cos  «  = p,    cos  /3  =  — —-,  cos  7  = 

Voo  VUG  Voo 

1  2  3 

cos  «  =  — — ,    cos  /3  =  -  — ,  cos  7  =  — — « 

Vl4  Vl4  Vl4 


60°. 


Page  217 


COS-1  -,     COS" 


6 


-,     COS 


■'!)■ 


4.   (fV3,  |,  0);   (|V3, 
6.    cos"1  £V15. 


o,  I). 


Page  220 

3.  y2  +  z2  =  25  ;  x2  +  z2  =  25. 

4.  a2  +  ?/2  +  z2  -  6  x  -  4  y  -  10  z  +  13  =  0. 

5.  7s  +  7y+10z  =  9;2a;-y-72=-30. 

Page  223 

2.  x2  +  z2  -  6  x  -  10  2  +  9  =  0. 

3.  (a)  y2  +  z2  =  16.  (c)  x2  +  y2  +  z2  =  r2. 
(6)  x2  -  y2  -  z2  =  0.        (cZ)  y2  +  z2  =  2  ma;. 

4.  &2x2  -  a2y2  -  a2z2  =  a2b2 ;  62x2  -  a2*/2  +  b2z  =  a2b2. 


Page  230 


1.  (a)    JL,    -=L*,    _1_.       (6)   J§_. 

V74      V74     V74  V74 

2.  3  x  +  y  -  5  2  =  35. 


,  .      19 

V74 


00  4,   - 

6.    3  x  +  59  y  —  72  z  =  0. 


Page  233 

1.  (a)  11 »  -  17  y  -  13  2  +  3  =  0.  (&)4s-7y-5s-£ 

2.  12  x-  17?/  +  42  -3  =  0.  5.  2se  +  z-8  =  0. 

3.  y  +  3  z  +  3  =  0.  6.    10  x  -  19  y  -  32  z  =  0. 

4.  7x  +  4y-4z-22=0. 


=  0. 


278  ANSWERS 

Page  236 

1.  x  =  y  =  0;   y  =  z  =  0;    z  =  x  =  0. 

2.  -s  =  0,  Ax  + 2ty  +  G  =  0,  etc.  ;  2  =  &,  ^4x  +  #y  +  C  =  0,  etc.  ; 

s  =  A*i,  ?/  =  &2,  etc. 

4.  (a)  x  =  $z  +  l  y  =  iz-i.  (d)  y  =  2,  z  =  -  3. 
(?))  a:  =  -■¥*,  y  =  ^2+4.  (e)  *  =  fy--l,  2  =  1. 
(c)  e  =  -  4,  a  =  -  2  y  +  3.                    (f)  y  =  -l,  z  =  2. 

5.  s  =  0,  «  =  f  y  +  3  ;    x  =  0,  z  =  3y  +  6;    y  =  0,z=-2x  +  fi. 

Page  238 

2.  (a)  re  =  */  -  3,  x  =  \\/2z  +  |V2  -  1. 
(ft)  se  =  -y+l,  x  =  \\'2  z  +  l\/~2  -  \. 
(r)  x  =  V3  y  -  2  V3  -  1,  z  =  -  3. 

3.  x  =  y  =  z. 

Page  240 

3-    (a)  ?,    =A   jj.  (c)  4=i   =1.   0. 


7  y/b      V5 

3'   3'   3* 


(&)  It   %   %  (<0  1>  0,  0 


Page  241 

3.  (a)  x  =  -3y,  (b)  x  =  f  z  -  4,  (c)  a;  =  2, 

e  =  -  2  y  -  2.  y  =  -  f  *  +  8.  y  =  -  3. 

4.  x  =  0,2  y  +  z  =  0.        8.    (4,1,-2).       9.    12 x  -  by  -  bz  -  31  = 

Page  243 

1.  (a)  x2  +  y*  +  z2  -  10  x  +  4  y  -  6  z  +  37  =  0. 
(6)  x'2  +  */2  +  22  -  4  x  +  6  y  +  12  z  =  0. 

(c)   x2  +  if-  +  22  ±  2  ax  =  0. 

2.  (a)   (1,-3,  4),  2.  (d)   (-3,  0,0),  3. 

(b)  (-5,  2,  1),  5.  (e)    Imaginary. 

(c)  (-2,  -2,  -3),  4. 

3.  (a)  Indeterminate  ;  points  lie  in  a  plane. 
(b)  x2  +  y2  +  z2-2x-8y-16z  =  0. 

4.  x2  +  y2  +  z*-2x-2y  -2z  =  0. 

5.  x2  +  y2  +  s2  =  4. 


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